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LIBRARY 

THE  UNIVERSITY 
OF  CALIFORNIA 

SANTA  BARBARA 

PRESENTED  BY 

Glen   G.    Mosher 


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GRAPHIC   STATICS 


WITH   APPLICATIONS   TO 


TRUSSES,  BEAMS,  AND  ARCHES. 


BY 

JEROME  SONDERICKER,  B.S.,  C.E., 

Associate   Professor   of  Applied   Mechanics, 
Massachusetts  Institute  of  Technology. 


FIRST   EDITION. 
FIRST   THOUSAND. 


NEW  YORK: 

JOHN  WILEY  &  SONS. 

LONDON:  CHAPMAN  &  HALL,  LIMITED. 

1904. 


Copyright,  1903, 

BY 
JEROME  SONDERICKER. 


ROBHRT  DRUMMOND,    PRINTER,    NEW  YORK. 


PREFACE. 


THIS  book  is  the  outgrowth  of  an  extended  experience  in 
teaching  Graphic  Statics  at  the  Massachusetts  Institute  of  Tech- 
nology. While  it  deals  specifically  with  problems  encountered 
in  building  construction,  it  should  be  found  serviceable  to  en- 
gineers and  engineering  students  generally. 

As  preparation,  the  reader  should  have  a  knowledge  of  Statics 
and  Strength  of  Materials,  including  beam  stresses  and  deflec- 
tions, as  these  subjects  are  commonly  presented. 

To  be  successful  in  the  employment  of  graphical  methods, 
it  is  necessary  not  only  to  understand  the  general  principles  in- 
volved, but  also  to  know  how  to  proceed  in  the  construction 
of  the  drawings,  in  order  to  secure  the  most  satisfactory  results. 
This  matter  is  given  more  attention  than  usual. 

An  attempt  has  been  made  in  §  6,  Chapter  II,  to  develop  a 
general  method  of  dealing  with  frames  where  bending  stresses 
occur  in  addition  to  the  tension  and  compression  stresses.  It  is 
hoped  this  may  be  found  useful. 

JEROME  SONDERICKER. 

MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY, 
May,  1903. 

at 


CONTENTS. 


CHAPTER   I. 

GENERAL  METHODS. 

§   I.       INTRODUCTION. 
ART.  FAGS 

I.  Definition  of  Graphic  Statics I 

a.   Representation  of  Forces I 

3.  Resultant  of  Forces  lying  in  the  Same  Plane I 

4.  Example „  3 

5.  Conditions  of  Equilibrium t 4 

6.  Examples  and  Problems 5 

§  2.      FUNICULAR   POLYGON. 

7.  Definitions 8 

8.  Applications IO 

9.  Notation • II 

10.  Problems 13 

11.  Distributed  Forces 14 

12.  Funicular  Polygon  for  a  Distributed  Force 14 

13.  Funicular  Polygon  for  a  Uniformly  Distributed  Force 15 

§  3.      FORCES   IN   EQUILIBRIUM. 

14.  Use  of  Funicular  Polygon 17 

15.  Case  I.     Parallel  Forces 17 

16.  Case  II.  Non-parallel  Forces 19 

17.  Case  III.   Non-parallel  Forces 20 

18.  Resolution  of  Forces  into  Components 21 

§  4.       GRAPHICAL   DETERMINATION    OF   MOMENTS. 

19.  Moment  of  Resultant  of  any  System  of  Forces 21 

20.  Moment  of  Resultant  of  Parallel  Forces 22 

v 


yi  CONTENTS. 

§  5.      SOME  SPECIAL   CONSTRUCTIONS   FOR   FUNICULAR   POLYGONS. 
ART.  PAGE 

21.  Locus  of  Points  of  Intersection  of  Corresponding  Strings 23 

22.  Locus  of  Poles  of  Funicular  Polygons  passing  through  Two  Given  Points.  24 

23.  Funicular  Polygon  through  Three  Points 25 

24.  Funicular  Polygon  through  Three  Points.     Parallel  Forces 25 

25.  Funicular   Polygon   through    Two   Points,  One    String    having   a  Given 

Direction 


CHAPTER   II. 

ROOF-TRUSSES. 

§  I.      CONSTRUCTION    OF   ROOFS;   LOADS,    ETC. 

26.  Construction  of  Roofs 30 

27.  Loads. . .  o . . . „! 

§  2.      DETERMINATION    OF   REACTIONS   OF  SUPPORTS. 

28.  Cases 33 

29.  Examples 35 

30.  Algebraic  Solution 37 

§     3.      DETERMINATION   OF  STRESSES. 

31.  General  Methods 38 

32.  Method  of  Sections 38 

33.  Method  of  Joints 39 

34.  Determination  of  Stresses  in  Roof-trusses 40 

35.  Example.     Bow's  Notation 40 

36.  Determination  of  Maximum  and  Minimum  Stresses 44 

37.  Example « 45 

38.  General  Remarks 49 

39.  Problems 5  r 

40.  Cantilever  Truss 53 

41.  Trusses  having  only  Two  Forces  to  determine  at  Each  Joint 53 

42.  Fink,  or  French,  Roof-truss 53 

§  4.      COUNTERBRACING  AND  DOUBLE   SYSTEMS   OF   BRACING. 

43.  Definitions 55 

.44.  Notation 56 

45.  Determination  of  Stresses.     Diagrams  drawn  for  Combined  Loads 56 

46.  Example 60 

47.  Determination  of  Stresses  in  Trusses  with  Counterbracing.   Second  Method  62 


CONTENTS.  vii 

ART.  PAGE 

48.  Trusses  having  a  Double  System  of  Web  Members 64 

49.  Double  Diagonal  Bracing 65 

§  5.       THREE-HINGED   ARCH. 

50.  Definition 66 

51.  Determination  of  Reactions  of  Hinges . .  66 

c  2.  Determination  of  Reactions.     Algebraic  Solution 68 

53.  Determination  of  Stresses  in  Braced  Arches 69 

54.  Three-hinged  Arch.     Solid  Ribs.     Determination  of  Stresses 72 

55.  Bending  Moments  proportional  to  Vertical  Intercepts 74 

§  6.      BENDING   STRESSES,    SWAY-BRACING. 

56.  Conditions  under  which  Bending  Stresses  occur 75 

57.  Purlins  supported  at  Other  Points  than  the  Joints 76 

58.  Trussed  Beam 79 

59.  Incomplete  Trusses 79 

60.  Trusses  supported  by  Columns 82 

6 1 .  Case  I.  Columns  hinged  at  Base 82 

62.  Case  II.  Columns  hinged  at  Base  and  loaded  only  at  the  Joints 87 

63.  Case  III.  Columns  fixed  at  Base 87 

64.  Case  IV.  Columns  fixed  at  Base  and  loaded  only  at  the  Joints 89 

65 .  Example 90 

66.  Approximate  Solution 92 

67.  Problems 94 

68.  Sway-bracing.     Illustration 95 

69.  Example 96 


CHAPTER   III. 


§  I.      SHEARING   FORCE   AND  BENDING  MOMENT. 

70.  Definitions 98 

71.  Graphical  Representation  of  Shearing  Force  and  Bending  Moment 98 

72.  Relation  between  Shear  and  Moment  Diagrams 100 

73.  Relation  between  Moment  Diagram  and  Elastic  Curve 101 

74.  Examples 101 

75.  Problems 106 


viii  CONTENTS. 

§  2.      DEFLECTION   OF  BEAMS. 

ART.  PAGE 

76.  Graphical  Determination  of  Elastic  Curve 107 

77.  Examples no 

78.  Problem 1 12 

79.  Center  of  Gravity  of  Any  Quadrilateral  Area 112 


CHAPTER   IV. 

MASONRY  ARCHES,  ABUTMENTS,   ETC. 
§  I.      GENERAL  CONDITIONS  OF  STABILITY. 

80.  Nature  of  the  Forces  involved 114 

81.  Resistance  of  a  Masonry  Joint 115 

82.  Resistance  to  Overturning 115 

83.  Resistance  to  Sliding 115 

84.  Resistance  to  Crushing 116 

85.  Conditions  to  be  satisfied  by  Masonry  Arches,  Abutments,  etc 117 

§  2.      MASONRY  ARCH.      LINE  OF   PRESSURE. 

86.  Definitions 118 

87.  Line  of  Pressure  a  Funicular  Polygon 118 

88.  A  Test  of  Stability 119 

89.  Relation  between  Line  of  Pressure  and  Form  of  Arch-ring 122 

90.  Maximum  and  Minimum  Crown  Pressure 124 

91.  Location  of  True  Line  of  Pressure 124 

92.  Example 129 

93.  Unsymmetrical  Cases 131 

§  3.      ABUTMENTS,    PIERS,    ETC. 

94.  Conditions  of  Stability 133 

95.  Example  1 133 

96.  Example  2 134 

97.  Example  3 135 

98.  Example  4 135 


GRAPHIC  STATICS. 


CHAPTER   I. 

GENERAL  METHODS. 

§  i.  Introduction. 

1.  Graphic    Statics    has  for  its  object  the  solution  of   prob- 
lems in  statics  by  means  of  geometrical  constructions,  the  results 
being  obtained  directly  from  the  scale  drawings. 

(A  knowledge  of  Statics  is  presupposed.  However,  a  brief 
presentation  of -its  principles  and  methods  in  case  of  forces  lying 
in  the  same  plane  will  be  made. 

In  the  subsequent  chapters,  familiarity  with  the  subject  of 
Strength  of  Materials,  including  beam  stresses  and  deflections,  is 
also  assumed.) 

2.  Representation  of    Forces.     A  force    is   fully  determined 
when  its  magnitude,  direction,  and  point  o]  application  are  known. 
In  dealing  with  problems  in  Statics  of  Rigid  Bodies,  the  magnitude, 
direction,  and  line  of  action  of  a  force  are  the  elements  commonly 
involved,  since  the  equilibrium  or  motion  of  such  a  body  is  not 
affected  by  transferring  the  point  of  application  of  a  force  to  any 
other  point  in  its  line  of  action. 

3.  Resultant    of    any    System     of    Forces     Lying     in    the 
Same  Plane.      The  magnitude  and  direction  of  the  resultant  of 


2  GRAPHIC  STATICS. 

any  system  of  forces  lying  in  the  same  plane  may  be  found  by 
either  of  the  following  methods: 

1.  Geometrically.    Represent  the  given  forces  by  the  sides  of 
a  polygon  taken  in  order.     The  closing  side  in  reverse  order  is 
the  resultant  in  magnitude  and  direction. 

2.  Algebraically.      Resolve   each    force  F  into    components 
Fx  and  FVj  parallel  to   coordinate  axes  X  and  Y  respectively. 
Then  RX=IFX  and  Ry=IFy.     Combining  these  we  have 


(i) 
(2) 


Equations  (i)  and  (2)  give  the  magnitude  and  direction  of  the 
resultant,  R  being  its  magnitude  and  ar  the  angle  it  makes  with  X. 

The  line  0}  action  of  the  resultant  may  be  found  by  either 
of  the  following  methods: 

la.  Geometrically.  Combine  the  forces  in  succession  by 
means  of  the  principle  that  the  resultant  of  two  forces  lying  in 
the  same  plane  must  pass  through  their  point  of  intersection. 

2a.  Algebraically.  Use  the  method  of  moments,  i.e.,  The 
moment  of  the  resultant  of  any  such  system  of  forces,  about  any 
axis  perpendicular  to  the  plane  of  the  forces,  is  equal  to  the  alge- 
braic sum  of  the  moments  of  the  forces. 

In  case  the  magnitude  of  the  resultant  is  zero,  the  forces  either 
form  a  couple  or  are  in  equilibrium.  If  the  resultant  is  a  couple, 
its  moment  can  be  found  by  2a  or  the  given  forces  can  be  combined 
into  a  single  resultant  couple  by  la.  Equilibrium  is  the  special 
case  of  a  couple  whose  moment  is  zero. 

If  the  lines  of  action  of  the  given  forces  intersect  at  a  common 
point,  the  line  of  action  of  the  resultant  will  pass  through  this  point, 
its  magnitude  and  direction  being  found  by  the  methods  pre- 
viously stated. 


GENERAL   METHODS, 


It  will  be  observed  that  the  determination  of  the  resultant, 
when  it  is  a  single  force  and  not  a  couple,  involves  two  opera- 
tions, namely:  i.  The  determination  of  its  magnitude  and  direc- 
tion. 2.  The  determination  of  its  line  of  action.  We  may  per- 
form both  of  these  operations  graphically  by  means  of  i  and  ia, 
or  algebraically  by  means  of  2  and  20. 

4.  Example.  Find  the  resultant  of  the  four  forces  F,  Ff,  F", 
F'"  (Fig.  i  A)  by  each  of  the  preceding  methods. 

First  Solution.  Represent  the  given  forces  by  the  sides  of  the 
polygon  (Fig.  i  B)  taken  in  order;  then  the  closing  side,  AD,  repre- 
sents the  magnitude 
and  direction  of  the 
resultant.  The  nu- 
merical values  of  these 
quantities  may  be 
found  (i)  by  solving 
the  polygon  by  trig- 
onometry, or  (2)  by 
direct  measurement  from  a  scale  drawing. 

To  find  the  line  of  action  of  the  resultant  by  the  first  method, 
we  can  proceed  as  follows:  The  magnitude  and  direction  of  the 
resultant  of  F  and  F'  is  AB.  Its  line  of  action,  R',  is  drawn 
parallel  to  AB  through  the  intersection  of  the  lines  of  action  of  F 
and  F'.  Continuing  in  the  same  manner,  we  determine  R"  to  be 
the  line  of  action  of  the  resultant  of  F,  F',  and  F",  and  finally 
R  to  be  the  line  of  action  of  the  resultant  of  the  four  forces. 

Second  Solution.  Resolve  each'  of  the  four  forces  into  hori- 
zontal (H)  and  vertical  (F)  components.  Then 

2H=\$o+ioo  cos  45°+ 0—150  cos  30°  =  90.8, 
2V  =  0+100  sin  45°+ 80+ 1 50  sin  30°  =  225. 7, 


(B) 


FIG.  i. 


GRAPHIC  STATICS. 


>+(2T)2=  243.3, 

/y  TT\ 
-,         _          l-K-)=68°5'. 

To  find  the  line  of  action  of  R,  apply  the  method  of  moments. 
Using  O  as  moment  axis,  we  have 

2M  =150-0—  ioo-  1.414+80-0—  1  50  -.866=  —  271.3. 
Hence  the  moment  of  the  resultant  about  O  is  left-handed,  and  its 
distance  from  O  is  -^-  =  1.115.     This  locates  the  resultant  as 

given  in  Fig.  i. 

The  moment  arms  of  the  several  forces  may  be  computed  or 
may  be  measured  directly  from  a  scale  drawing  when  the  results 
thus  obtained  are  sufficiently  exact. 

PROBLEM.  Represent  four  forces  by  the  sides  of  a  closed  poly- 
gon taken  in  order.  Assume  the  lines  of  action  of  these  forces  at 
random.  Find  the  moment  of  the  resultant  couple  by  each  of  the 
two  methods  of  Art.  3. 

5.  Conditions  of  Equilibrium.  Forces  not  acting  at  the  same 
point.  The  conditions  of  equilibrium  may  be  deduced  from 
the  fact  that  any  system  of  balanced  forces  can  be  reduced  to  two 
equal  and  opposite  forces  having  the  same  lines  of  action,  i.e.,  a 
couple  whose  moment  is  zero. 

A.  If  the  resultant  is  a  couple,  R  =  o,  hence: 

1.  Geometrically.     The   given   forces   can  be  represented   in 
magnitude  and  direction  by  the  sides  of  a  closed  polygon  taken 
in  order. 

2.  Algebraically.  If  the  forces  be  resolved  into  components 
parallel  to  coordinate  axes,  the  algebraic  sum  of  each  set  of  com- 
ponents must  be  zero,  i.e.,  IFx  =  o  and  IFV  =  <3. 


GENERAL  METHODS.  5 

B.  In  order  for  the  moment  of  the  resultant  couple  to  be  zero, 
we  have: 

i  a.  Geometrically.  The  line  of  action  of  the  resultant  of  any 
portion  of  the  given  forces  must  coincide  with  the  line  of  action 
of  the  resultant  of  the  remainder. 

2a.  Algebraically.  The  algebraic  sum  of  the  moments  of  the 
forces  must  be  zero. 

i  and  la  constitute  the  geometrical,  and  2  and  2a  the  algebraic 
conditions  of  equilibrium. 

Forces  acting  at  the  same  point.  In  this  case  condition  i  or  2 
is  sufficient.  The  condition  that  if  three  non-parallel  forces 
balance  they  must  intersect  at  a  common  point,  is  a  special  case 
under  la. 

When  any  system  of  forces  lying  in  the  same  plane  is  in  equilib- 
rium, one  or  more  of  the  preceding  conditions  of  equilibrium  serve 
to  determine  the  unknown  elements  of  the  problem,  if  it  is  solvable 
under  the  assumption  that  the  body  acted  on  is  rigid. 

6.  Examples,  i.  The  truss  (Fig.  2)  is  in  equilibrium  under 
the  supporting  forces  P  and  Pr  and  the  load  W.  P  is  vertical 
and  W,  the  known  force,  is  normal  to  the  roof  surface  at  its  middle 
point  O.  Indicate  how  to  find  P  and  P'  by  each  of  the  following 
methods  (see  Art.  5):  (i)  By  using  la  and  i;  (2)  by  using  2a 
alone;  (3)  by  using  ia  to  find  one  force,  then  i  to  find  the  remain- 
ing force;  (4)  by  using  2a  to  find  one  force,  then  2  to  find  the 
remaining  force. 

First  Solution.  The  resultant  of  P  and  W,  acting  through 
N,  must  balance  the  remaining  force,  P',  acting  at  M;  hence  the 
line  of  action  of  P'  must  be  MN.  The  forces  P,  W,  and  Pf  to 
balance  must  form  a  triangle  (not  shown"),  the  sides  taken  in  order. 
Plotting  W  and  completing  the  triangle,  P  and  P'  are  determined. 

Second  Solution.     Substitute  for  P',  unknown  in  magnitude 


6  GRAPHIC  STATICS 

and  direction,  the  two  component  forces  PfH  and  P'v,  unknown 
in  magnitude  only.  Also,  for  convenience,  substitute  for  W  its 
H  and  V  components.  To  find  P  take  the  moment  axis  at  M. 
the  intersection  of  the  two  other  unknown  forces.  Then  the 
algebraic  sum  of  the  moments  of  P,  WH,  Wv,  P'H  and  P'v  about 
M  must  be  zero.  Solving  the  equation  thus  formed,  we  determine 


'    : so- 


FIG.  2. 


P.  Similarly,  to  find  Py,  take  moments  about  Of.  To  find 
P'H  take  moments  about  any  convenient  point  not  on  its  line  of 
action,  e.g.,  O. 

Third  Solution.  Find  one  force,  e.g.,  P,  by  moments,  then 
plotting  P  and  W,  the  closing  side  of  the  triangle  formed  gives 
the  magnitude  and  direction  of  P'. 

Fourth  Solution.  Find  one  force,  as  P,  by  moments.  Then. 
since  2FX  and  IFy  must  each  be  zero,  we  have 

P'  =  W      and     P'V  =  WV-P. 


and       tan  a  =      - 


EXAMPLE    2.     The  portion  of  the  truss  (Fig.  2)  to  the  left  of 
AB  is  in  equilibrium  under  the  action  of  the  supporting  force  P, 


GENERAL   METHODS.  7 

load  W,  and  the  forces  F,  Ff,  and  F"  exerted  by  the  portion  of  the 
truss  to  the  right  of  the  section  AB  upon  the  left-hand  portion. 
The  lines  of  action  of  these  latter  forces  coincide  with  the  centre 
lines  of  the  members  cut  by  AB,  and  their  magnitudes  are  equal 
to  the  stresses  existing  in  these  members.  P  and  W  being  known, 
indicate  how  to  find  the  three  unknown  forces  by  each  of  the 
following  methods  (see  Art.  5) : 

(i)  By  using  conditions  la  and  i;  (2)  by  using  20.  alone;  (3) 
by  using  20  to  find  one  force,  then  i  to  find  the  remaining  two; 
(4)  by  using  20-  to  find  one  force,  then  2  to  find  the  remaining 
two  forces. 

First  Solution.  The  resultant  of  P  and  F"  acting  through  Or 
must  balance  the  resultant  of  W,  F,  and  F'  acting  through  O.  " 

_       *1 

Hence  the  line  of  action  of  each  resultant,  R,  must  be  OO'. "  P, 
F",  and  their  resultant  R,  must  form  a  triangle  (Fig.  2  A) ;  P 
being  the  known  force,  F"  and  R  are  thus  determined.  The  re- 
sultant, R,  must  balance  the  forces  at  O,  hence  R,  W,  F,  and 
F'  must  form  a  closed  polygon  as  shown  in  Fig.  2  A.  F  and  F' ', 
the  remaining  unknown  forces,  are  thus  determined  by  plotting 
W  and  completing  the  polygon. 

Second  Solution.  To  find  jP,  take  the  moment  axis  at  the 
intersection  O"  of  the  other  two  unknown  forces,  so  that  their 
moments  will  each  be  zero.  Then  the  algebraic  sum  of  the 
moments  of  P,  W,  and  F  about  O"  must  equal  zero.  Solving 
the  equation  thus  formed,  we  determine  F.  Similarly,  to  find 
F'  take  moments  about  O',  and  to  find  F"  take  moments  about  O. 

Third  Solution.  We  find  one  force,  as  F,  by  the  preceding 
method,  then  represent  the  known  forces  P,  W,  F,  by  the  sides  of 
a  polygon  taken  in  order,  and  complete  the  polygon  by  lines 
parallel  to  the  two  remaining  forces,  Ff  and  F". 

Fourth  Solution.  We  find  one  force,  F,  by  the  method  of  mo- 
ments as  before.  Then,  placing  the  algebraic  sums  of  the  hori- 


GRAPHIC  STATICS. 

zontal  and  vertical  components  of  the  forces  each  equal  to  zero, 
we  form  two  equations  which  are  solved  for  the  two  remaining 
unknown  forces,  F'  and  F". 

In  algebraic  solutions  the  directions  of  the  unknown  forces 
are  assumed,  the  algebraic  signs  of  the  results  indicating  whether 
the  assumed  directions  are  correct  or  not. 

PROBLEM  i.  In  Example  i,  obtain  the  numerical  results, 
solving  by  each  of  the  four  methods  mentioned.  W  =  5000  Ibs. 

PROBLEM  2.  In  Example  2,  obtain  the  numerical  results, 
solving  by  each  of  the  four  methods  mentioned.  Also  determine 
the  kind  of  stress  (tension  or  compression)  in  the  members  cut 
by  AB. 

Remarks.  The  student  should  become  familiar  with  each 
mode  of  solution  so  as  to  be  able  to  select  readily  that  one  which 
is  best  adapted  to  the  problem  at  hand. 

Ability  to  make  such  selection  can  be  secured  only  by  solving 
a  variety  of  problems  by  various  methods  and  comparing  the 
merits  of  the  different  solutions. 

In  using  the  method  of  moments,  it  should  be  noted  that  it  is 
usually  better  to  use  the  H  and  V  components  of  the  forces  than 
it  is  to  use  the  original  forces. 

It  should  also  be  noted  that  in  case  of  forces  not  acting  at  the 
same  point  but  lying  in  the  same  plane,  three  unknown  quantities 
can  be  determined,  while  if  the  forces  act  at  the  same  point  only 
two  can  be  determined. 


§  2.  Funicular  Polygon. 

7.  Definitions.  Let  F,  F',  F"  (Fig.  3  A),  be  given  forces,  their 
magnitudes  being  represented  by  AB,  BC,  CD  (Fig.  3  B).  Assume 
any  point  P,  and  draw  the  radial  lines  PA,  PB,  etc.  From  any 
point  M  on  the  line  of  action  of  F  draw  ML  and  MN  parallel 


GENERAL   METHODS. 


(8) 


FIG.  3. 


to  PA  and  PB  respectively.  From  N,  where  MN  intersects  the 
line  of  action  of  F',  draw 
NO  parallel  to  PC;  simi- 
larly draw  OQ  parallel  to 
PD,  thus  forming  the  broken 
lineLMNOQ.  We  may  con- L 
sider  AP  and  PB,  having 
the  directions  of  the  arrows 
marked  i,  to  be  the  com- 
ponents of  the  force  F,  the 
lines  of  action  of  these  p 
components  being  ML  and 
MN  respectively.  Similarly, 
BP  and  PC,  having  the 
directions  marked  2  and  the 
lines  of  action  NM  and  NO,  may  be  taken  as  components  of 
F' ;  and  CP  and  PD  with  ON  and  OQ  for  lines  of  action  as  the 
components  of  F" '.  MN  is  thus  the  line  of  action  of  two  equal 
and  opposite  forces  PB,  and  NO  of  the  two  equal  and  opposite 
forces  PC.  These  two  pairs  of  forces  consequently  balance, 
leaving  APand  PD,  having  ML  and  OQ  for  their  lines  of  action, 
as  the  equivalent  of  the  original  forces. 

The  broken  line  LMNOQ  is  called  a  funicular  or  equilibrium 
polygon.  The  former  name  is  given  because  the  line  corresponds 
to  the  shape  assumed  by  a  weightless  cord  when  fastened  at  the 
ends  and  acted  on  by  the  given  forces.  This  is  shown  by  the 
polygon  drawn  in  dotted  lines.  The  latter  name  is  applicable 
since  a  jointed  frame  of  the  form  of  the  polygon  would  be  in  equi- 
librium under  the  action  of  the  given  forces. 

The  point  P  is  called  the  pole;  the  lines  PA,  PB,  etc.,  are  the 
rays;  and  the  corresponding  lines  of  the  funicular  polygon  are 
its  strings. 


10  GRAPHIC  STATICS. 

Figure  3  A  is  called  the  space  diagram,  since  it  represents  the 
location  of  the  lines  of  action  of  the  forces.  Its  scale  is  one  of 
distance.  Figure  3  B  is  the  force  diagram,  the  lengths  of  its  lines 
representing  the  magnitudes  of  the  forces  to  scale.  The  perpen- 
dicular distance  from  the  pole  to  any  side  of  the  force  polygon  is 
called  the  pole  distance  of  that  force.  It  is  to  be  noted  that  this 
distance  represents  a  force  magnitude. 

In  the  case  of  parallel  forces  the  force  polygon  becomes  a 
straight  line,  and  the  pole  distances  of  all  the  forces  are  equal. 

8.  Applications.  The  following  results  are  readily  derived 
from  the  construction  explained  in  Art.  7. 

(1)  The  resultant  of  F,  F',  and  F"  is  given  in  magnitude  and 
direction  by  the  closing  side  AD  of  the  force  polygon,  and  its  line 
of  action  passes  through  the  point  of  intersection  of  the  strings 
LM  and  OQ.     The  resultant  is  thus  completely  determined;  and, 

L  in  general,  the  line  of  action  of  the  resultant  0}  any  system  o)  forces, 
\taken  consecutively,  passes  through  the  point  0}  intersection  o]  the 
,  two  strings  between  which  the  forces  lie. 

(2)  If  the  force  polygon  is  closed,  PA  and  PD  will  coincide,  and 
hence  the  corresponding  strings  LM  and  OQ  will  be  parallel     In 
this  case  the  resultant  is  a  couple  whose  arm  is  the  perpendicular 
distance  between  the  parallel  strings,  and  whose  forces  are  repre- 
sented by  the  ray  (PA  =PD)  corresponding  to  these  strings. 

(3)  In  order  for  the  given  forces  to  be  in  equilibrium,  the  arm 
of  the  couple  in  (2)  must  be  zero;  that  is,  the  strings  LM  and  OQ 
must  coincide  in  MO.     In  this  case,  the  funicular  polygon  is  said 
to  be  closed.     The  conditions  of  equilibrium,  therefore,  are  that 
both  the  force  and  funicular  polygons  must  close. 

If  the  forces  intersect  at  a  common  point,  the  first  of  these 
conditions  is  sufficient,  since  such  a  set  of  forces  cannot  form  a 
couple. 


GENERAL  METHODS.  1 1 

(4)  Any  number  of  funicular  polygons  may  be  drawn  for  the 
same  system  of  forces  by  using  different  poles  and  beginning  the 
construction  of  the  polygons  at  different  points.  Various  geomet- 
rical relations  exist  between  these  different  polygons,  some  of  which 
are  given  in  §  5.  The  following  relations  are  derived  directly 
from  the  preceding  discussion. 

(a)  Corresponding  pairs  of  non-parallel  strings  of  the  various 
funicular  polygons  must  intersect  on  the  same  straight  line,  this 
being  the  line  of  action  of  the  resultant  of  the  forces  included  by 
these  strings. 

(b)  In   the  case  where    the   force   polygon   closes,    by  using 
different  poles  different  couples  will  be  obtained;   but  as   these 
couples  are  all  equally  the  resultant  of  the  given  system  of  forces, 
their  moments  will  be  equal. 

It  is  obvious  that  any  string  can  be  given  any  desired  direction 
by  drawing  the  corresponding  ray  in  this  direction  and  assuming 
the  pole  at  any  point  on  the  line  thus  drawn. 

In  selecting  the  pole,  the  obtaining  of  accurate  and  convenient 
diagrams  is  kept  in  view.  Generally  the  rays  should  not  make 
very  oblique  angles  with  the  adjacent  lines  of  the  force  polygon.  If 
the  pole  is  taken  at  a  vertex  of  the  force  polygon,  as  at  A,  Fig.  3  B, 
the  construction  of  the  funicular  polygon  becomes  identical  with 
that  explained  in  Art.  4,  for  finding  the  line  of  action  of  the  resultant 
by  id-  The  construction  in  Art.  4  frequently  leads  to  inaccu- 
rate and  inconvenient  diagrams  and  is  inapplicable  to  parallel 
forces. 

9.  Notation.  Resultant  of  a  System  of  Forces.  To  illustrate 
the  notation  to  be  used,  let  it  be  required  to  find  the  resultant  of  the 
four  forces  AB,  BC,  CD,  DE  (Fig.  4).  The  line  of  action  of  any 
force  is  designated  by  the  two  letters  between  which  it  lies ;  thus, 
ab  represents  the  line  of  action  of  the  first  force,  be  of  the  second 


GRAPHIC  STATICS. 


force,  etc.     In  the  force  diagram,  the  same  letters  in  capital  type 
are  used,  but  placed  at  the  ends  of  the  lines  representing  the 

forces.  In  constructing  the 
force  polygon,  the  forces  and 
letters  must  be  taken  in  order 
from  the  space  diagram  (e.g., 
proceeding  from  left  to  right)." 
When  this  is  done,  if  the  letters 
are  read  in  order  from  left  to 
right  in  the  space  diagram,  the 
same  order  of  letters  in  the  force 
polygon  is  used  to  indicate  the 
directions  in  which  the  successive 
forces  act.  It  will  be  noticed  that 


FIG.  4. 


the  string  corresponding  to  the  ray  PA  lies  in  the  space  a,  that 
corresponding  to  PB  in  the  space  b,  etc.  These  strings  will  be 
referred  to  as  the  strings  a,  b,  etc.,  and  the  corresponding  rays  as 
the  rays  A ,  B,  etc. 

Having  constructed  the  force  and  funicular  polygons,  the 
magnitude  and  direction  of  the-  resultant  is  represented  by  the 
closing  side,  AE,  of  the  force  polygon,  and  one  point  in  its  line 
of  action  is  the  point  of  intersection  of  the  similarly  lettered  strings 
(a  and  e).  The  line  of  action  of  the  resultant,  R,  is  then  drawn 
through  this  point  of  intersection,  parallel  to  AE.  In  referring  to 
any  force  (e.g.,  the  force  BC),  the  capital  letters  will  be  employed, 
although  the  line  of  action  of  the  force  as  well  as  its  magnitude 
and  direction  is  included  in  the  reference. 

The  system  of  notation  just  described  is  applicable  in  most 
cases  occurring  in  engineering  practice.  When  the  location  of 
the  forces  is  such  that  it  cannot  be  used  to  advantage,  the 
modified  notation  illustrated  in  Fig.  5  may  be  employed  instead. 
This  needs  no  description. 


GENERAL   METHODS.  ij 

In  general,  in  referring  to  forces  in  the  text,  the  letters  will 
be  used  so  as  to  indicate  by  their  order  of 
succession  the  direction  in  which  the  forces 
act. 

Special  Case.  In  finding  the  resultant 
of  a  system  of  forces  the  case  may  occur 
in  which  the  two  strings  whose  intersection 
locates  the  resultant  are  parallel  (but  do 
not  form  a  couple),  or  are  so  nearly  parallel  A 
that  they  do  not  intersect  conveniently.  In 
such  cases  the  resultant  may  be  located  as  follows:  Let  it  be 
required  to  find  the  resultant  of  the  forces  BC,  CD,  and  DB 
(Fig.  4).  This  resultant  is  BE,  its  line  of  action  passing  through 
the  intersection  of  the  strings  b  and  e,  which  here  are  parallel. 
Now  we  know  (Art.  7)  that  the  forces  under  consideration  are 
equivalent  to  the  two  forces  represented  by  the  end  strings  b  and  e 
and  the  corresponding  rays,  BP  and  PE.  Hence  we  can  solve  the 
problem  by  finding  the  resultant  of  these  two  forces  by  means  of 
a  second  funicular  polygon,  as  follows :  Taking  the  point  A  as  a, 
convenient  pole,  draw  a  funicular  polygon  (shown  in  dotted  lines) 
for  the  forces  BP  and  PE.  The  intersection  of  the  end  strings 
at  Rr  locates  the  resultant  which  is  drawn  parallel  to  BE. 

-^— 
10.  Problems. 

(1)  Assume  five  non-parallel  forces,  and  find  their  resultant, 
using  two  different  poles. 

(2)  Assume    five    non-parallel    forces    whose    force    polygon 
closes,  and  find  their  resultant,  using  two  different  poles. 

(3)  Assume  five  parallel  forces  and  find  their  resultant,  using 
two  different  poles. 

(4)  Assume  five  parallel  forces  whose  algebraic  sum  is  zero 
and  find  their  resultant,  using  two  different  poles.  t 


14  GRAPHIC  STATICS, 

11.  Distributed   Forces.      Distributed    forces    are  commonly 
represented  in  terms  of  their  intensity.     The  intensity  of  a  dis- 
tributed force  is  the  force  per  unit  length,  area,  or  volume,  as  the 
case  may  be.     Gravity  and  water  pressure  are  familiar  illustra- 
tions of  such  forces. 

If  the  intensity  is  the  same  at  all  points,  the  force  is  unijormly 
distributed. 

Parallel  forces  lying  in  the  same  plane  will  alone  be  consid- 
ered here.  Such  forces  are  repre- 
sented as  follows: 

Let  AB  (Fig.  6)  be  any  line  per- 
pendicular to  the  direction  of  the 
forces,  dx  any  infinitesimal  portion 
of  AB,  and  let  the  ordinate,  length 

p,  represent  the  intensity  of  the  force  at  the  point  from  which  the 
ordinate  is  drawn.  The  shaded  area  pdx  thus  represents  the 
amount  of  force  acting  on  the  length  dx  of  the  line.  The  diagram 
(Fig.  6)  is  then  to  be  interpreted  as  follows: 

The  length  of  the  ordinate  at  any  point  of  the  line  AB  repre- 
sents the  intensity  of  the  force  at  that  point,  and  the  area  included 
between  any  two  ordinates  represents  the  amount  of  force  acting 
on  the  corresponding  portion  of  AB. 

12.  Funicular  Polygon  for  a  Distributed  Force.     Let  WXYZ 
(Fig.  7)  represent  a  distributed  force.     Divide  XZ  into  equal 
divisions  sufficiently  small  so  that  the  force  represented  by  any 
area,  as  STUV,  may  be  taken  to  act  at  the  middle  point  n  of  the 
division  and  the  area  may  be  taken  to  be  equal  to  UVXnm. 
Then  the  distributed  force  is  equivalent  to  the  four  forces  AB, 
BC,  CD,  and  DE,  their  magnitudes  (AB,  BC,  etc.,  of  the  force 
polygon)  being  proportional  to  the  lengths  of  the  corresponding 
middle  ordinates,  such  as  mn.    The  funicular  polygon  GH  is 


GENERAL   METHODS.  15 

drawn  from  this  system  of  forces.  Now  it  is  evident  that  as  the 
divisions  of  XZ  are  made  smaller,  this  polygon  would  more 
closely  approach  the  true  funicular  polygon  (curve)  for  the  dis- 
tributed force.  The  true  funicular  polygon  would  then  be  a  curve 
tangent  to  the  polygon  GH  at  the  points  (marked  with  short 
lines)  corresponding  to  the  points  of  division  of  the  distributed 


FIG.  7. 


force.     Each  string  would  be  infinitesimal  in  length  and  have 
the  direction  of  a  tangent  to  the  curve. 

Such  a  funicular  polygon  can  be  used  for  the  same  purposes 
and  in  a  similar  manner  to  one  drawn  for  concentrated  forces. 
Thus  to  find  the  line  of  action,  R,  of  the  resultant  force  AE,  we 
would  draw  tangents,  GK  and  HK,  to  the  curve  at  the  points 
corresponding  to  X  and  Z,  these  tangents  being  drawn  parallel  to 
the  rays  A  and  E  of  the  force  diagram. 


13.  Funicular  Polygon  for  a  Uniformly  Distributed  Force. 

A  familiar  construction  of  analytical  geometry  for  a  parabola 
with  axis  vertical  is  shown  in  Fig.  8.  It  consists  in  dividing  a 
vertical  line  1-8  into  equal  parts  and  drawing  a  series  of  equidistant 
vertical  lines  as  shown.  The  broken  line  i'-8'  is  then  obtained 
by  drawing  its  successive  segments,  i',  2',  etc.,  parallel  to  Pi, 


i6 


GRAPHIC  STATICS. 


P2,  etc.     A  curve  tangent  to  this  broken  line  at  the  middle  points 
of  its  segments  is  a  parabola  with  axis  vertical. 

Comparing  this  construction  with  that  of  Fig.  7,  we  see  that  it 
is  identical  with  the  construction  of  a  funicular  polygon  for  a 


FIG.  8. 


uniformly  distributed  force.  The  funicular  polygon  for  a  uni- 
formly distributed  jorce  is  thus  a  parabola  whose  axis  is  parallel 
to  the  force. 

If,  then,  the  force  WXYZ  (Fig.  7)  were  uniformly  distributed, 


B' 


FIG.  9. 


the  funicular  curve  would  be  a  parabola,  axis  vertical,  tangent 
to    the  strings   KG  and   KH   at   the   points   corresponding   to 


GENERAL   METHODS.  1 7 

the  limits  of  the  force.     Such  a  parabola  can  be  constructed  as 
follows  (see  Fig.  9) : 

Plot  the  resultant  force  AB,  its  line  of  action  R  bisecting  XZ, 
and  draw  the  strings  Kx  and  Kz  parallel  to  the  rays  A  and  B 
respectively.  Divide  Kx  and  Kz  into  the  same  number  of  equal 
parts,  as  at  i,  2,  3  and  i',  2'  and  3'.  Lines  joining  these  points  of 
division,  as  shown,  are  tangents  to  the  required  curve.  The 
points  of  tangency,  indicated  by  circles,  are  located  by  drawing 
lines  parallel  to  R  through  the  alternate  points  of  division  (com- 
pare with  Fig.  8). 

§  3.  Forces  in  Equilibrium. 

14.  Use  of  Funicular  Polygon.     It  will  be  noticed  from  §  2 
that  the  funicular  polygon  construction  is   especially  adapted  to 
the  case  of  parallel  and  other  non- concurrent  forces,  although  it 
may  be  used  to  advantage  in  problems  relating  to  forces   inter- 
secting at  a  common  point,  when  this  point  lies  outside  the  limits 
of  the  drawing.     The  conditions  of  equilibrium,  given  in  Art.  8, 
for  non- concurrent  forces  are  (i)  that  the  force  polygon   must 
close,  and  (2)  that  the  funicular  polygon  must  close.     The  appli- 
cation of  these  conditions  to  the  solution  of  problems  will  now  be 
illustrated,  ,the  cases  most  frequently  arising  in   practice  being 
presented. 

15.  Case   I.     Parallel   Forces.      Given  a    system  of  parallel 
forces  in  equilibrium,  the  lines  of  action  of  all  and  the  magnitudes 
and  directions  of  all  but  two  being  known.     It  is  required  to  find 
the  unknown  elements. 

Let  the  unknown  forces  be  the  supporting  forces  of  the  beam 
(Fig.  10).  Represent  the  loads  in  succession  by  the  sides  AB, 
BC,  CD  of  the  force  polygon,  and,  selecting  a  suitable  pole,  draw 


1 8  GRAPHIC  STATICS. 

the  strings  a,  b,  c,  d  of  the  funicular  polygon.  The  string  a 
intersects  the  left  reaction  at  w,  and  the  string  d  the  right  reac- 
tion at  n. 

The  line  joining  m,  n  is  then  the  closing  string  e  of  the  funicular 
polygon,  and  the  ray  PE  drawn  parallel  to  it  determines  DE  and 

EA  to  be  the  right-  and  left- 

B  -    a    ijt^t*^?     I  hand  supporting  forces  re- 

E    I   ^  ^""^^TT  sPectiyety-     When  the  un- 

c    \m    e  -•$    known  (supporting)  forces 

are  parallel,  but  the  given 
FlG- I0-  forces  (loads)  are  not,  the 

supporting  forces  must  be  parallel  to  the  resultant  load.  If  we 
take  AD  (Fig.  10)  to  represent  this  resultant  load,  the  construc- 
tion for  determining  the  magnitudes  of  the  supporting  forces  will' 
be  similar  to  that  just  explained. 

The  method  of  determining  the  magnitudes  and  directions  of 
the  supporting  forces  from  the  lettering,  after  the  construction  is 
completed,  should  be  carefully  noted.  It  is  as  follows: 

(1)  The  strings  are  lettered  with  the  same  letters  as  the  cor- 
responding rays,  so  that  the  strings  intersecting  on  the   line  of 
action  of  any  force  have  the  same  letters  as  those  which  represent 
that  force  in  the  force  polygon.     The  two  strings  intersecting  on 
the  left  supporting  force  are  a  and  e,  hence  AE  represents  the 
magnitude  of  this  force.     Similarly  d  and  e  intersect  on  the  right 
supporting  force;  so  its  magnitude  is  DE. 

(2)  The  forces  are  laid  off  in  the  force  polygon  in  right-handed 
order;  i.e.,  if  the  letters  representing  them  in  the  space  diagram 
be  read  in  right-handed  order,  the  same  order  of  letters   in  the 
force  polygon  indicates  the  direction  in  which  the  forces  act.     We 
also  know  that  the  force  polygon  must  close.     The  force  polygon 
then  is  ABCDEA,  the  order  of  letters  DE,  EA    indicating  that 
each  supporting  force  acts  upwards.     It  should  be  noted  in  this 


GENERAL   METHODS.  19 

and  the  following  solutions  that  the  known  forces  are  made  to  follow 
consecutively  in  constructing  the  force  polygon. 

(To  obtain  a  thorough  understanding  of  these  solutions,  it 
would  be  well  for  the  student  to  trace  out  the  construction  from 
the  standpoint  of  the  triangle  of  forces.  Thus,  in  the  present  case, 
the  force  AB  is  resolved  into  AP  and  PB,  whose  lines  of  action 
are  the  strings  a  and  b.  Similarly  BC  is  resolved  into  BP  and 
PC,  and  CD  into  CP  and  PD.  The  two  forces  PB  and  BP 
balance  since  they  are  equal  and  opposite  and  have  the  same 
line  of  action,  b.  PC  and  CP  balance  for  a  similar  reason;  hence 
the  three  loads  are  equivalent  to  AP  and  PD,  whose  lines  of  action 
are  a  and  d  respectively.  Now,  in  order  to  have  equilibrium, 
the  resultant  of  AP  and  the  left  supporting  force  must  balance 
the  resultant  of  PD  and  the  right  supporting  force.  The  lines  of 
action  of  these  two  resultants  must  then  coincide  in  mn.  Since, 
of  the  three  forces  intersecting  at  m,  one  is  the  resultant  of  the 
other  two,  they  must  form  a  triangle,  AP  being  the  known  magni- 
tude. This  triangle  is  APE;  thus  the  reaction  EA  is  deter- 
mined. The  corresponding  triangle  for  the  forces  intersecting 
in  n  is  EPD.) 

16.  Case  II.  Non-parallel  Forces.  Given  a  system  of  forces 
in  equilibrium,  all  but  two  of  which  are  known  completely.  Of 
these  two,  the  line  of  action  of  one  and  one  point  in  the  line  of 
action  of  the  other  are  known.  It  is  required  to  determine  the 
unknown  elements. 

Let  the  known  forces  be  the  wind  pressures  AB,  BC,  CD  on  the 
roof  (Fig.  n).  The  unknown  forces  are  the  reactions  of  the  sup- 
ports. The  line  of  action  of  the  right  supporting  force  is  given, 
it  being  vertical;  and  one  point,  m,  of  the  left  reaction  is  known. 
Represent  the  loads  in  succession  by  the  sides  AB,  BC,  CD  of 
the  force  polygon.  The  strings  e  and  a  must  intersect  on  the 


GRAPHIC  STATICS, 


line  of  action  of  the  supporting  force  EA,  and  as  the  point 
m  of  this  line  of  action  is  known,  the  funicular  polygon  will  be 
constructed  so  that  these  strings  will  intersect  at  this  point.  The 
construction  of  the  funicular  polygon  is  then  begun  by  drawing 


FIG.  ii. 

the  string  a  through  the  point  m,  the  strings  b,  c,  d  being  then 
drawn  in  order. 

The  string  d  intersects  the  right  supporting  force  at  n,  and,  of 
course,  the  string  a  intersects  the  left  supporting  force  at  m;  mn 
is  thus  the  closing  side,  e,  of  the  funicular  polygon.  The  ray  cor- 
responding to  the  string  e  .is  now  drawn,  intersecting  the  right 
supporting  force,  DE,  at  E.  The  triangle  of  forces,  PDE,  whose 
lines  of  action  intersect  at  n,  is  thus  formed,  and  the  magnitude 
of  the  supporting  force  DE  is  determined.  Finally,  the  closing 
side,  EA,  of  the  force  polygon  represents  the  other  supporting 
force  in  magnitude  and  direction.  Its  line  of  action  is  drawn 
through  m,  parallel  to  EA. 

17.    Case  III.     Non-parallel    Forces.       Given    a    system    of 
forces  in  equilibrium,  the  lines  of  action  of  all  are  known,  but  the 
magnitudes  and  directions  of  three  of  the  forces 
are  unknown.     It  is  required  to  determine  the  un- 
known elements.     Let  ab,  be,  cd  (Fig    12)  be  the 
FIG.  12.         lines  of  action  of  the  three  unknown  forces.     The 
resultant  of  any  two,  as  be  and  cd,  must  pass  through  their  point 


GENERAL   METHODS.  21 

of  intersection,  n.  If  this  resultant  be  substituted  for  be  and  cd, 
the  problem  becomes  identical  with  Case  II;  ab  being  the  line  of 
action  of  one  unknown  force,  and  n  one  point  of  the  line  of  action 
of  the  other. 

The  construction  of  Case  II  having  been  made,  the  resultant 
of  be  and  cd  can  be  resolved  into  components  parallel  to  these  lines 
of  action,  thus  completing  the  solution.  Figure  13  shows  this 


FIG.  13. 

method  applied  to  determine  the  stresses  in  the  members  ef,  jg, 
and  ga,  of  the  truss,  the  loads  and  reactions  being  known. 


18.  Resolution  of  Forces  into  Components.  Since  forces 
equal  and  opposite  to  the  components  and  having  the  same  lines 
of  action  will  balance  the  given  forces,  we  can  solve  problems  of 
this  nature  in  the  same  manner  as  if  the  forces  balanced,  the 
desired  components  being  the  balancing  forces  with  their  direc- 
tions reversed. 


§  4.  Graphical  Determination  of  Moments. 

19.  Moment  of  Resultant  of  any  System   of  Forces.    The 
resultant  of  AB,  EC  and  CD  (Fig.  14)  is  AD,  its  line  of  action, 


GRAPHIC  ST/tTICS. 


FIG.  14. 


lettered    R,  passing    through    th^    intersection    of    the    strings 

a  and  d.  The  moment 
8  of  this  resultant  about 
any  point  M  is  thus 
(R  =  AD)x,  in  which  x  is 
the  moment  arm.  Now 
draw  S T  through  M  par- 
allel to  AD  and  termina- 
ting in  the  strings  a  and  d, 
thus  forming  with  these  strings  a  triangle  (drawn  in  heavy  lines) 
similar  to  PAD.  From  these  two  similar  triangles  we  have, 
AD-x  =  ST-y.  But  AD-x  is  the  moment  of  the  resultant 
force,  hence 

The  moment  of  the  resultant  of  any  system  0}  forces  may  be 
found  by  drawing  a  line  parallel  to  the  resultant  through  the  moment 
axis,  noting  the  distance  intercepted  on  this  line  by  the  strings  cor- 
responding to  the  resultant.  The  product  of  this  intercept  into  the 
pole  distance  of  the  resultant  is  the  desired  moment.  This  method 
has  no  value  except  in  case  of  parallel  forces. 

20.  Moment  of  Resultant  of  Parallel  Forces.  The  con- 
structions of  this  section  are  especially  adapted  to  the  case  of 
parallel  forces.  Illustration:  Let  the 
beam  (Fig.  15)  be  loaded  as  shown. 
The  reactions  of  the  supports  (Art.  15) 
are  found  to  be  CD  and  DA.  It  is  re- 
quired to  find  the  bending  moment  at 
any  section,  ST.  By  definition,  this 
bending  moment  is  equal  to  the  algebraic  ° 
sum  of  the  moments  of  the  forces  to  the 
left  of  the  section,  and,  by  Art.  19,  this  c 
resultant  moment  is  equal  to  the  pole 
distance,  y,  times  the  intercept  ST. 


FIG,  15. 


GENERAL   METHODS. 


Now  in  case  of  parallel  forces,  the  pole  distance  is  the 
same  for  all  the  forces,  hence  the  bending  moments  at  the 
various  sections  of  the  beam  are  directly  proportional  to  the 
corresponding  intercepts.  This  subject  is  further  developed  in 
Chapter  III. 

It  follows  directly  from  Arts.  19  and  20  that  if  any  two  funicular 
polygons  be  constructed  for  a  given  system  of  parallel  forces,  the 
ratio  of  the  intercepts  of  corresponding  pairs  of  strings  on  any  line 
drawn  parallel  to  the  forces  is  constantj  being  equal  to  the  inverse 
ratio  of  the  pole  distances  of  the  respective  polygons. 

§  5.   Some  Special  Constructions  for  Funicular  Polygons. 

21.  Locus  of  Points  of  Intersection  of  Corresponding  Strings* 

JA  //  any  two  funicular  polygons  be  drawn  for  the  same  system  of  forces, 
])  their  corresponding  strings  will  intersect  on  a  straight  line  parallel 
I  to  the  line  joining  the  two  poles. 

Proof.  Let  AB,  BC,  CD  (Fig.  16)  be  the  given  forces.  (The 
student  should  keep  both  diagrams  in  mind,  also  that  the  order 
of  letters  in  the  text  indicates  the  directions  of  the  forces.)  The 
funicular  polygons  for 
the  two  poles,  P  and 
Pf,  are  drawn  in  full 
and  broken  lines  re- 
spectively. 

The  resultant  of 
PA  and  AP'  is  PPf. 
Its  line  of  action  xz 
is  parallel  to  PP'  and 
passes  through  x,  the  point  of  intersection  of  the  strings  a  and  a' 
which  are  the  lines  of  action  of  the  forces  PA  and  AP'  respectively. 
Similarly,  the  resultant  of  PB  and  BP'  is  PP',  its  line  of  action 
yz  passing  through  the  intersection  of  the  strings  b  and  V,  etc. 


FIG.  16. 


24  GRAPHIC  STATICS, 

Now  PB  is  the  resultant  of  PA  and  AB,  also  BPf  is  the  result- 
ant of  BA  and  AP'.  Since  ^4£  and  BA  are  balanced  forces,  the 
resultant  of  PB  and  J3P'  is  the  same  as  the  resultant  of  PA  and 
AP' ;  hence  their  lines  of  action,  xz  and  yz,  must  coincide.  The 
same  reasoning  applies  to  the  intersections  of  the  remaining  strings. 
The  points  x,  y,  etc.,  therefore  lie  on  the  line  xz,  parallel  to  PP' ', 
this  being  the  line  of  action  of  the  common  resultant  of  the  pairs 
of  forces,  PA  and  AP' \  PB  and  BP'\  PC  and  CP',  etc. 

22.  Locus  of  Poles  of  Funicular  Polygons  passing  through 
Two  Given  Points.  The  locus  of  the  poles  0}  all  funicular  poly- 
gons, two  of  whose  corresponding  strings  pass  through  two  given 
points,  is  a  straight  line  parallel  to  the  line  joining  the  two  points. 

Proof.  Let  AB,  BC,  CD  (Fig.  17)  be  the  given  forces  and 
x  and  y  be  the  points  through  which  the  strings  a  and  d  respectively 


FIG.  17. 

are  to  pass.  Assume  two  balancing  forces  DE  and  EA,  parallel 
to  AD  and  passing  through  the  points  y  and  x  respectively.  The 
magnitudes  of  these  balancing  forces  are  found,  as  in  Art.  15,  by 
drawing  the  closing  string  e  (xfyf)  of  the  funicular  polygon  and 
the  corresponding  ray  E.  Now  since  the  forces  ABCDEA  are 
balanced,  all  funicular  polygons  drawn  for  them  will  close.  If, 
then,  we  wish  the  string  a  of  any  such  polygon  to  pass  through  x 
and  the  string  d  through  y,  the  string  e  must  evidently  pass  through 
both  x  and  y,  and  the  pole  P'  therefore  lie  somewhere  on  the  ray 
drawn  through  E  parallel  to  xy,  i.e.,  parallel  to  the  line  joining  the 
two  given  points. 


GENERAL   METHODS.  2$ 

23.  Funicular    Polygon    through  Three   Points.     PROBLEM. 

To  draw  a  funicular  polygon  for  a  given  system  of  forces,  such  that 
three  designated  strings  shall  pass  through  three  given  points. 
Let  the  forces  be  ABCDEF  (Fig.  18).  It  is  required  to  draw  a 
funicular  polygon  such  that  the  string  a  will  pass  through  O,  the 
string  c  through  O',  and  the  string  /  through  O". 


FIG.  18. 

Let  mn  be  any  funicular  polygon  for  the  given  forces,  with  P 
for  pole.  By  means  of  the  construction  of  Art.  22,  we  determine 
P'X  to  be  the  locus  of  the  poles  of  all  funicular  polygons  whose 
strings  a  and  c  pass  through  O  and  O'  respectively.  Also,  by  the 
same  construction  we  determine  P'X'  to  be  the  locus  of  the  poles  of 
all  funicular  polygons  whose  strings  c  and  /  pass  through  O'  and 
O"  respectively.  Hence,  in  order  for  both  conditions  to  be  satis- 
fied, the  pole  must  lie  on  both  P'X  and  P'X',  i.e.,  at  their  point 
of  intersection,  P'.  The  required  polygon  is  drawn  in  full  lines. 

(Note.  To  secure  accuracy,  draw  the  strings  a,  c,  and  /  first, 
then  draw  the  intermediate  strings  closing  on  the  ones  midway 
between  the  given  points.) 

24.  Funicular  Polygon  through  Three  Points.     Parallel  Forces. 

A  shorter  solution  than  that  given  in  Art.  23  can  be  made  for  this 
case.    Let  AB,  BC,  CD,  and  DE  (Fig.  19)  be  the  given  forces,  it 


26 


GRAPHIC  STATICS. 


being  required  to  draw  a  funicular  polygon  for  them  such  that 
the  string  a  will  pass  through  O,  the  string  c  through  O',  and  the 
string  e  through  O".  Let  mn  be  any  funicular  polygon  for  the 
given  forces,  P  being  the  pole.  Draw  lines  parallel  to  the  forces 
through  each  of  the  three  given  points.  Taking  the  ones  through 
O  and  O"  to  be  the  lines  of  action  of  balancing  forces,  and  follow- 
ing the  construction  of  Art.  22,  we  determine  mn  to  be  the  closing 
string  (/)  of  the  polygon  drawn,  the  closing  string  of  the  required 


FIG.  19. 

polygon  being  OO".  Knowing  that  the  string  c  of  the  required 
polygon  must  pass  through  O',  the  intercept  on  a  line  through 
O'  made  by  the  strings  c  and  /  of  this  polygon  will  be  O'S',  the 
corresponding  intercepts  of  the  first  polygon  being  RS.  Now, 

O'S' 
the  ratio  -^-F  being  constant  for  all  corresponding  intercepts  of 

the  two  polygons  (see  Art.  20),  the  vertices  of  the  required  poly- 
gon may  be  determined  as  follows:  Lay  off  R"S"  =  RS  (Fig. 
19  A).  From  S"  describe  an  arc,  radius  S"X  =  O'S'  and  draw  the 
tangent  R"T.  Then  to  determine  any  intercept,  for  example 
uv,  take  the  distance  u'i/  in  the  dividers  and  step  it  off  from  R" 
on  R"S" .  The  perpendicular  distance  from  the  point  thus  located 


GENERAL   METHODS. 


27 


to  R"T  is  the  desired  intercept.     The  polygon  drawn  in  full  lines 
was  thus  locaied. 

In  case  the  intercept  RS  is  shorter  than  O'S',  Fig.  19  A  must 
be  modified  by  making  R"S"  =  n-RS,  where  n  =  2,  3,  etc.  In 
such  case  the  proportion  will  be 

uv  :O'S'  ::n-u'i/:n-RS. 

The  pole  P'  of  the  required  polygon  can  be  located  on  P'F 
(see  Art.  22)  by  means  of  the  proportion 

y-.y.'.RS-.O'S'  (see  Art.  20). 

PROBLEM.  Draw  a  funicular  curve  (parabola)  for  a  uniformly 
distributed  load  so  as  to  pass  through  three  given  points  (Fig.  20). 


FIG.  20. 

Let  the  load  be  distributed  over  the  length  OO",  the  three 
points  being  O,  O',  and  O" .  Draw  lines  (vertical)  parallel  to 
the  load  through  these  three  points.  Substitute  for  the  portions 
of  the  load  lying  on  the  two  sides  of  the  vertical  through  Or, 
their  resultants  AB  and  BC,  and  construct  any  funicular  polygon 
mn  for  them,  P  being  the  pole.  The  corresponding  funicular  curve 
would  be  tangent  to  the  strings  a,  b,  and  c  at  m,  R,  and  n  respect- 


GRAPHIC  STATICS. 


ively  (see  Art.  13).  A  second  funicular  polygon  whose  string  b 
will  pass  through  O'  is  now  drawn,  the  vertices  i  and  2  being 
located  as  explained  for  Fig.  19.  The  strings  Oi,  1-2,  and  zO" 
of  this  polygon  will  be  tangent  to  the  required  funicular  curve  at 
O,  O',  and  O"  respectively.  This  curve  is  constructed  by  the 
method  shown  in  Fig.  9. 

25.  Funicular  Polygon  through  Two  Points,  One  String  having 
a  Given  Direction  PROBLEM.  To  draw  a  funicular  polygon 
for  a  given  system  of  forces,  such  that  two  designated  strings  shall 
pass  through  two  given  points  and  one  string  of  the  polygon  shall 
have  a  given  direction. 

The  method  of  Art.  22  can  be  used  to  determine  the  locus  of 
the  poles  of  all  polygons  passing  through  the  two  given  points. 
The  intersection  of  the  ray  corresponding  to  the  string  whose 
direction  is  given  with  this  locus  will  be  the  pole  of  the  required 
polygon. 

If  the  string  whose  direction  is  given  is  also  one  of  the  two 
which  are  to  pass  through  the  given  points,  the  following  simpler 
solution  can  be  made.  Let  AB,  BC,  and  CD  (Fig.  21)  be  the 

given  forces,  and  P  the 
pole  of  any  funicular  poly- 
gon mn  for  these  forces. 
It  is  required  to  draw  a 
second  polygon  such  that 
the  strings  a  and  d  will 
pass  through  O  and  O'  re- 
spec  a vely,  and  the  string 
a  be  horizontal.  The  re- 
FlG  2J  sultant  of  the  forces  in- 

cluded between  the  desig- 
nated strings  a  and  d  is  AD,  its  line  of  action  passing  through  the 


GENERAL  METHODS  29 

intersection  S  of  the  strings  a  and  d  of  the  first  polygon.  The 
corresponding  strings  of  the  second  polygon  must  intersect  on 
this  line  (Art.  8).  These  strings  are,  therefore,  OS',  drawn 
horizontally,  and  S'O'.  The  new  pole  P'  will  be  at  the  inter- 
section of  the  corresponding  rays.  The  polygon  can  now  be 
completed. 


CHAPTER  II. 


ROOF-TRUSSES. 

§  i.  Construction  of  Roofs,  Loads,  etc. 

26.  Construction  of  Roofs.  A  roof  includes  the  covering 
.and  the  framework.  The  covering  is  tin,  slate,  tarred  felt,  cor- 
rugated iron,  etc.,  laid  over  sheathing;  or  the  sheathing  may  be 
omitted,  as  is  commonly  done  when  corrugated  iron  is  used. 

The  roof  covering  is  supported  by  a  more  or  less   elaborate 
framework,  the  principal  members  of  which  are    often  trusses 
•or  arches.     A  common  arrangement  of  this  framework,  in  case  of 
^JACK-RAFTER         shops,  train-sheds,   assembly    halls, 
etc.,  is  shown  in  plan  and  elevation 
in  Fig.  22      The   trusses,  AB,   are 
made  up  of  an  upper  chord,  a  lower 
£  chord,    and    the   web    members.     At 
right  angles  to  the  trusses  and  sup- 
ported   by    them    are    a    system    of 
members,  CD,  called  purlins.     The 
purlins  are  usually  wooden  or  iron 
beams,  but  in  case  of  wide  spacing 
between   the   trusses   they   may   be 
FlG-  *2-  trussed  members.     It  is  desirable  to 

support  the  purlins  at  the  joints  of  the  upper  chord,  otherwise 
bending  stresses  would  occur  in  these  members.  Sometimes  the 
covering  is  laid  directly  on  the  purlins,  but  more  commonly  jack- 
raj  ters,  EF,  supported  by  the  purlins,  serve  to  further  subdivide 

30 


'     ROOF-TRUSSES.  31 

the  area,  the  covering  being  laid  on  them.  A  system  of  sway- 
bracing  is  employed,  when  necessary,  to  keep  the  trusses  vertical 
and  in  line.  Such  a  system  is  represented  in  Fig.  22  by  the  diag- 
onal rods  S,  S,  situated  in  the  plane  of  the  upper  chord. 

Roof-trusses  are  supported  either  by  masonry  walls  or  piers, 
or  by  wooden  or  iron  columns.  When  iron  trusses  of  considerable 
span  rest  on  masonry,  a  more  or  less  effectual  provision  for  expan- 
sion consists  in  supporting  one  end  on  a  planed  bed-plate  or  on 
rollers. 

(For  a  range  of  temperature  of  120°  F.  iron  changes  in  length 
about  i  inch  in  100  ft.)  To  provide  for  expansion  lengthwise 
of  an  iron  framed  building,  expansion  joints  are  inserted  in  the 
longitudinal  members  at  suitable  intervals.1 

27.  Loads.  The  loads  which  have  commonly  to  be  provided 
for  in  designing  a  roof-truss  are: 

(1)  Dead  Load.     This  includes  the  weight  of  the  covering, 
framework,  and  any  other  permanent  loads  supported  by  the 
truss,  such  as  the  weights  of  floors,  ceilings,  etc.     This  loading 
varies  from  about  8  Ibs.  to  35  Ibs.  or  more  per  square  foot  of  roof 
surface. 

(2)  Snow  Load.     This  is  estimated  at  10  Ibs.  to  25  Ibs.  per 
horizontal  square  foot  in  northern  United  States. 

(3)  Wind  Load.     This  is  estimated  at  30  Ibs.  to  50  Ibs.  per 
square  foot  on  a  vertical  surface,  the  resulting  normal  pressure 

1  For  detailed  descriptions  of  roofs,  see  the  following  references: 

"Revue  Technique  de  1'Exposition  de  Chicago,"  Part  I,  Architecture  (with 
Atlas).  E.  Bernard  &  Co.,  Paris. 

"Philadelphia  and  Reading  Terminal  Station,  Philadelphia."  Trans.  Am. 
Soc.  C.  E.,  1895. 

"The  South  Terminal  Station,  Boston,  Mass."  Trans.  Am.  Soc.  C.  E.,  Dec., 
1899. 

"St.  Louis  Coliseum."     Jour.  Assoc-  Eng.  Soc.,  May,  1898. 

"Dome  of  Government  Building,  World's  Columbian  Exposition."  Trans. 
Am.  Soc.  C.  E.,  Jan.,  1892. 


GRAPHIC  STATICS. 


on  an  inclined  surface  being  determined  by  some  empirical 
formula.1 

Provision  should  be  made  for  any  additional  loads  to  be  sup- 
ported by  the  roof,  e.g.,  live  loads  on  floors,  weights  of  shafting 
and  travelling  cranes,  pull  of  belts,  thrust  of  jib-cranes,  etc. 

The  pressures  at  the  various  joints  of  a  truss,  due  to  the  load- 
ing, are  determined  by  the  same  methods  as  the  supporting  forces 
in  case  of  a  beam.  The  direct  stresses  (tension  or  compression  - 
in  the  members  of  a  truss  depend  on  these  joint  pressures  or 
loads.  When  loads  are  applied  to  a  truss  at  other  points  than 
the  joints,  the  members  of  the  truss  thus  loaded  are  subjected 

1  Read  Lanza's  "Applied  Mechanics,"  §§  130-133,  from  which  the  following 
extract  is  taken: 

"Duchemin's  formula,  which  Professor  W.  C.  Unwin  recommends,  is  as  fol- 
lows, viz.: 

.     2  sin  6 


where  p=  intensity  of  normal  pressure  on  roof,  pt=  intensity  of  pressure  on  a  plane 
normal  to  the  direction  of  the  wind. 
"i)  Mutton's  formula, 

P=Pi  (sin  0)i-  84  cose-  1. 

V 

"Unwin  claims  that  this  and  Duchemin's  formula  give  nearly  the  same  results 
for  all  angles  of  inclination  greater  than  15°. 

"The  following  table  gives  the  results  obtained  by  the  use  of  each,  on  the 
assumption  that  />1=4o." 


e 

Duchemin. 

Hutton. 

e 

Duchemin. 

Hutton. 

5° 

6.89 

5-1° 

50° 

38.64 

38.10 

10° 

13-59 

9.60 

55° 

39-21 

39-40 

i5° 

19.32 

14.20 

60° 

39-74 

40.00 

20° 

24.24 

18.40 

65° 

39.82 

40.00 

25° 

28.77 

22.60 

7o° 

39-91 

40.00 

30° 

32.00 

26.50 

75° 

39-96 

40.00 

35° 

34-52 

30.10 

80° 

40.00 

40.00 

40° 

36.40 

33-3° 

85° 

40.00 

40.00 

45° 

37-73 

36.00 

90° 

40.00 

40.00 

ROOF-TRUSSES.  33 

to  bending  stresses  in  addition  to  the  direct  stresses.     This  case  is 
discussed  in  §  6. 

The  direct  stresses  in  the  members  of  a  roof-truss  can  usually 
be  determined  most  readily  by  graphical  methods.  In  order  to 
use  such  methods  to  the  best  advantage,  however,  it  is  necessary 
to  freely  employ  algebraic  methods  in  connection  with  them,  as 
indicated  in  the  following  pages. 

§  2.  Determination  of  Reactions  of  Supports. 

28.  Cases.  Let  the  loads  acting  on  the  truss  (Fig.  23)  be 
resolved  into  horizontal  (H)  and  vertical  (V)  components,  2H. 
and  2V  representing  the  algebraic  sums  of  the  horizontal  and1 
vertical  component  loads 
respectively,  and  IM  the  |V  * 

algebraic   sum  of  the  mo-  H^\' 

ments    of   the   loads.     Let  / 


HI,  Vu  and  H2,  F2  be  the       I    Vl  v,  "* 

components  of  the  left  and  Fro-  23- 

right  reactions  respectively.     Then  we  have 


IH,    .......     (i) 

V,+  F2=2T;    .......    (2) 

and.  Faking  moments  about  the  right  support, 

VJ-2M,    ........    (3) 

in  which  I  =  span.  . 

These  three  equations  involve  four  unknown  quantities. 
From  equations  (2)  and  (3)  V^  and  Vy  can  be  determined;  their 
values  are  thus  independent  of  the  manner  of  supporting  the  ends 
of  the  truss.  The  values  of  Hl  and  H2,  however,  vary  with  the 
manner  of  supporting  the  truss.  The  following  cases  occur: 


34  GRAPHIC  STATICS. 

CASE  I.  Truss  fixed  in  position  at  both  supports  by  anchor- 
bolts  or  otherwise.  When  the  roof  is  comparatively  flat,  so  that 
the  resultant  load,  including  wind  pressure,  is  nearly  vertical,  the 
reactions  can  be  assumed  parallel  to  each  other,  and  consequently 
parallel  to  the  resultant  load.  In  the  case  of  steep  roofs,  or  in 
general  where  the  resultant  load  makes  a  large  angle  with  the 
vertical,  this  assumption  may  lead  to  absurd  results. 

The  assumption  which  probably  approaches  the  truth  as 
closely  as  can  be  expected  for  so  indeterminate  a  case  is  that  the 
horizontal  reactions  o]  the  two  supports  are  equal.  If  the  truss  is 
assumed  to  be  rigid,  the  supports  equally  elastic,  and  no  other 
forces  besides  the  horizontal  pressure  on  the  truss  in  action  to 
cause  the  supports  to  yield,  this  assumption  would  be  correct; 
for,  since  the  supports  yield  equally,  the  horizontal  forces  causing 
such  yielding  must  be  equal. 

When  the  truss  is  not  anchored  to  its  supports,  the  conditions 
are  evidently  similar  to  Case  I,  so  long  as  the  weight  and  friction 
are  sufficient  to  prevent  motion. 

CASE  II.  One  end  of  truss^sup  ported  onjrollers^  planed  bed- 
plate, or  similar  device  to  provide  for  expansion.  If  friction  be 
neglected,  the  reaction  at  the  free  end  is  evidently  vertical.  Fric- 
tion at  the  free  end  may  be  dealt  with  as  follows :  First  determine 
the  supporting  forces  on  the  assumption  that  both  ends  are  fixed. 
If  the  resulting  reaction  at  the  free  end  makes  an  angle  with  the 
vertical  less  than  the  angle  of  repose,  these  reactions  are  the 
true  ones.  If,  however,  the  reaction  at  the  free  end,  thus  deter- 
mined, makes  a  greater  angle  with  the  vertical  than  the  angle 
of  repose,  both  reactions  must  be  determined  anew  on  the 
assumption  that  the  reaction  at  the  free  end  makes  an  angle  with 
the  vertical  equal  to  the  angle  of  repose. 

CASE  III.  Truss  supported  on  columns.  This  case  is  dis- 
cussed in  §  6. 


ROOF-TRUSSES. 


35 


29.  Examples.  The  methods  of  determining  the  supporting 
forces  are  included  under  Arts.  5,  15,  and  16.  A  few  examples 
are  given  to  indicate  suitable  solutions. 

EXAMPLE  i.  The  truss  (Fig.  24)  is  subjected  to  wind  pressure 
from  the  left,  the  load  being  uniformly  distributed  over  the  rafter. 
Find  reactions  of  supports:  (i)  When  both  ends  are  fixed  (Fig. 
24  A);  (2)  When  the  right  end  is  supported  on  rollers  (Fig.  24  B); 
(3)  When  the  left  end  is  supported  on  rollers  (Fig.  24  C). 

The  resultant  load  acts  at 
the  middle  of  the  rafter.  In 
Case  (i)  the  reactions  are  as- 
sumed to  be  parallel.  They 
are  found,  by  the  method  of 
Art.  15,  also  by  dividing  the 
resultant  load,  AD,  into  parts 
inversely  proportional  to  the 
segments  into  which  the  line 
of  action  of  the  resultant  load 
divides  the  line  joining  the 
supports. 

The  method  of  Art.  16  is 
used  to  solve  Cases  (2)  and 
(3),  the  reaction  at  the  free 
end  being  assumed  vertical. 
The  line  of  action  of  the  re- 


FIG.  24. 

action  at  the  fixed  end  is  also  determined  from  the  condition 
that  three  non-parallel  forces  in  equilibrium  must  intersect  at  the 
same  point,  after  which  the  magnitudes  of  the  reactions  can  be 
determined  by  the  triangle  of  forces. 

The  vertical  components  of  the  reactions  are  the  same  for  all 
three  cases  (see  Art.  28);   hence  the  results  for  any  one  of  these 


36  GRAPHIC  STATICS. 

cases  might  have  been  used  in  solving  the  other  two  cases  (see 
Example  2). 

EXAMPLE  2.  In  the  truss  (Fig.  25)  AB  and  BC  are  the  re- 
sultant wind  loads  on  the  vertical  and  inclined  surfaces  respect- 
ively, and  CD  is  the  resultant  dead  load.  It  is  required  to  deter- 
mine the  supporting  forces,  making  use  in  turn  of  each  of  the 
following  assumptions: 

(i)  Left  end  supported  on  rollers,  reaction  vertical;  (2)  left 
end  supported  on  smooth  bed-plate,  coefficient  of  friction  =  J;  (3) 
both  ends  fixed,  reactions  parallel;  (4)  both  ends  fixed,  horizontal 


FIG.  25. 

reactions  equal;    (5)  right  end  supported  on  rollers,  reaction  ver- 
tical. 

Solution.  The  line  of  action  of  the  resultant  of  BC  and  CD 
is  xy,  drawn  through  x  parallel  to  BD.  This  resultant  combined 
with  AB  gives  zy  for  the  line  of  action  of  the  resultant  load  AD. 
The  supporting  forces  under  assumption  i,  determined  by  the 
second  method  used  in  Example  i  for  Cases  2  and  3,  are  DE 
and  EA. 


ROOF-TRUSSES. 


37 


Now,  since  the  vertical  component  reactions  are  independent 
of  any  assumption,  the  point  corresponding  to  E  must  lie  on  a 
horizontal  line  through  E  in  all  cases.  Hence  the  remaining 
cases  can  be  solved  as  follows: 

Assumption  2.  Lay  off  EE'  =  %EA,  then  DE  and  E'A  are 
the  supporting  forces. 

Assumption  3.     DE"  and  E"A  are  the  supporting  forces. 

Assumption  4.  Bisect  EEIV  at  £'",  then  DE"  and  E"A 
are  the  supporting  forces. 

Assumption  5.     DEIV  and  EIVA  are  the  supporting  forces. 

It  should  be  noted  that  if  the  supporting  forces  had  been  de- 
termined for  the  dead  and  wind  loads  separately  and  the  results 
combined,  the  resultant  reactions  thus  found  would  differ  from 
those  in  Example  2  when  assumptions  2  and  3  are  employed, 
but  not  in  the  other  cases. 


r 


FIG.  26. 


PROBLEM.  The  truss  (Fig.  26)  is  subjected  to  wind  pressure 
on  the  left  side  as  shown,  in  addition  to  a  uniformly  distributed 
vertical  load  of  75,000  Ibs.  Determine  the  supporting  forces  under 
each  of  the  assumptions  of  Example  2. 

30.  Algebraic  Solution.  The  student  should  be  able  to 
determine  reactions  algebraically  as  well  as  graphically.  The 


GRAPHIC  STATICS. 


general  method  is  indicated  in  Art.  28.  Equations  (2)  and  (3) 
(Art.  28)  serve  to  determine  Vl  and  V2.  In  order  to  determine 
HI  and  H2,  we  must  have,  in  addition  to  Equation  (i),  a  second 
equation  based  on  the  assumption  employed.  Taking  the  five 
assumptions  of  Example  2  in  order,  this  second  equation  is  as 
follows: 


Assumption  i. 
2. 


H,    H2    IH 


r  I          V  2       ^  r 

«  .        TT  __  TT 

«  f  TT     

PROBLEM.     Solve  the  problem  (Fig.  26)  algebraically. 

§  3.  Determination  of  Stresses. 

31.  General  Methods.     A  truss  is  designed  to  support  loads 
applied  at  the  joints  by  virtue  of  the  resistance  to  extension  and 
compression  of  its  various  members. 

There  are  two  general  methods  for  determining  the  tension 
and  compression  stresses  in  the  members  of  a  truss:  (i)  method 
of  sections;  (2)  method  of  joints. 

32.  Method  of  Sections.     Let  the  imaginary  line  xy  (Fig.  27) 
divide  the  truss  into  two  parts,  this  line  intersecting  the  three 

members  dj,  jk,  ka.  The 
portion  of  the  truss  to  the 
left  of  xy  is  a  body  in  equi- 
librium under  the  action 
of  certain  forces.  Consider 
the  member  dj.  If  it  is  in 


FIG.  27. 


tension,  the  portion  to  the 


right   of  xy  must   be  exerting    force  upon  the  portion  to  the 


ROOF-TRUSSES.  39 

left  of  xy,  toward  the  right.  If  dj  is  in  compression,  its  right- 
hand  portion  must,  on  the  other  hand,  be  exerting  force  upon  its 
left-hand  portion,  towards  the  left.  In  either  case,  the  magnitude 
of  this  force  is  equal  to  that  of  the  stress  in  the  member,  and  the 
line  of  action  of  the  force  has  the  direction  of  the  length  of  the 
member.  This  force  is  external  with  reference  to  the  portion  of 
the  truss  to  the  left  of  xy.  It  will  be  represented  by  the  letters 
dj.  Similar  explanations  hold  for  jk  and  ka.  The  forces,  there- 
fore, which  hold  the  left  portion  of  the  truss  in  equilibrium  are  the 
known  forces  ab,  be,  cd,  and  the  forces  dj,  jk,  and  ka,  exerted  by 
the  right-hand  portions  of  these  three  members  upon  their  left- 
hand  portions.  The  lines  of  action  of  these  last  three  forces 
are  known,  their  magnitudes  and  directions  being  unknown. 
These  six  forces  constitute  a  system  of  forces  in  equilibrium,  lying 
in  the  same  plane,  but  not  acting  at  the  same  point.  One  or  more 
of  the  conditions  of  equilibrium  of  Art.  5  can  therefore  be  used  to 
determine  the  unknown  forces,  as  was  indicated  in  Art.  6.  The 
third  solution  of  Art.  6  (Example  2)  is  generally  most  useful  in 
dealing  with  roof-trusses. 

The  following  points  should  be  noted : 

(1)  Only  three  forces,  unknown  in  magnitude,  can  be  deter- 
mined, so  that  if  the  section  xy  cuts  more  than  three  members 
which  are  in  action  under  the  given  loads,  a  solution  cannot  be 
made. 

(2)  The  stress  in  a  member  is  equal  to  the  magnitude  of  the 
force  which  it  exerts,  and  the  nature  of  the  stress,  tension,  or 
compression  can  be  determined  from  the  direction  of  the  force 
(tension,   if  directed  away  from  left  portion;    otherwise  com- 
pression). 

33.  Method  of  Joints.     The  external  forces  acting  upon  the 
joint  b  (Fig.  27)  are:    (i)  the  supporting  force  ab;    (2)  the  load 


4°  GRAPHIC  STATICS, 

be;  (3)  the  forces  exerted  by  the  members  ck  and  ka  upon  the 
joint.  If  either  member  as  ck,  is  in  tension,  the  force  which  it 
exerts  on  the  joint  b  is  evidently  directed  away  from  the  joint; 
if  in  compression,  towards  it.  As  the  joint  is  in  equilibrium, 
these  four  forces  must  balance,  and  by  applying  either  the  alge- 
braic method  of  resolution  of  forces  or  the  geometric  method  of 
polygon  of  forces,  the  unknown  forces  can  be  determined. 

In  using  the  method  of  joints,  the  following  points  should  be 
noted : 

(1)  The  forces  dealt  with  are  those  acting  on  the  joint. 

(2)  In  dealing  with  any  one  joint  only  two  unknown  forces 
can  be  determined. 

(3)  The  nature  of  the  stress  in  a  member  can  be  determined 
from  the  direction  of  the  force  which  the  member  exerts  on  the 
joint:  tension,  if  the  force  acts  away  from  the  joint;  compression, 
if  the  force  acts  towards  the  joint. 

(4)  When  the  stress  in  a  member  is  determined,  the  force  which 
it  exerts  upon  the  joint  at  each  end  is  known,  these  forces  being 
equal  and  opposite. 

34.  Determination    of   Stresses    in   Roof-trusses.      Of    the 
methods  which  have  been  explained,  the  one  best  adapted  to  roof- 
trusses  is  the  method  of  joints,  solving  by  the  polygon  of  forces. 
In  applying  this  method,  the  external  forces  being  known,  we 
begin  by  constructing  the  polygon  of  forces  at  any  joint  of  the 
truss  where  only  two  stresses  are  unknown.     Having  thus  deter- 
mined these  two,  we  repeat  the  construction  for  another  joint 
where  only  two  stresses  remain  unknown,  and  continue  in  this 
manner  until  the  stresses  in  all  the  members  have  been  determined. 

35.  Example.     Bow's  Notation.     The  truss  (Fig.  28)  is  sub- 
jected to  a  uniformly  distributed  load,  W.     Each  intermediate 


ROOF-TRUSSES.  41 

joint  of  the  upper  chord  supports  \W,  and  each  end  joint  f  W.  Each 
supporting  force  is  ^W.  The  external 
forces  are  lettered  as  in  the  preceding 
chapter,  ab  representing  the  left  reac- 
tion; be,  cd,  etc.,  the  successive  loads. 
Letters  are  also  placed  in  the  spaces 
into  which  the  surface  is  divided  by 
the  web  members.  Each  member  of 
the  truss  is  represented  by  the  letters 
in  the  adjacent  spaces;  e.g.,  the  two 

halves    of    the    lower    chord    are    ka 

FIG.  28. 
and   ha;     the  vertical    member  is   ;?, 

etc.  The  directions  of  the  forces  are  indicated  in  the  force 
(or  stress)  diagram  by  the  order  of  the  letters.  For  this  purpose 
the  letters  are  to  be  read  in  right-handed  order  around  the  truss 
or  any  joint  of  it;  e.g.,  the  left  reaction  is  ab,  left-hand  load 
be  etc.;  the  force  exerted  by  the  member  ck  upon  the  left-hand 
joint  is  ck,  and  upon  the  joint  cdjk  is  kc. 

To  illustrate  the  manner  of  using  this  notation,  the  stresses  in 
the  various  members  of  the  truss  will  be  determined.  First  con- 
struct the  polygon  of  external  forces.  Lay  off  the  loads  on  a 
vertical  line  in  right-handed  order.  The  first  load,  beginning  at 
the  left,  is  be;  it  acts  downwards,  and  to  indicate  this  the  force  is 
lettered  so  that  BC  reads  downwards.  Continuing  in  this  manner, 
when  the  polygon  is  completed,  if  we  read  the  letters  surrounding 
the  truss  in  right-handed  order,  bcdefgab,  these  letters  in  the  stress 
diagram,  read  in  the  same  order,  will  represent  the  polygon  of 
external  forces,  the  order  of  succession  of  the  letters  indicating 
the  direction  in  which  the  forces  act. 

Next  the  polygon  of  forces  for  the  left-hand  joint  of  the  truss 
is  constructed.  The  known  forces  ab  and  be  are  already  included 
in  the  stress  diagram.  The  two  unknown  forces  are  ck  and  ka. 


42  GRAPHIC  STATICS. 

One  letter  of  each  (C  and  A)  is  already  in  the  stress  diagram. 
These  letters  indicate  the  points  from  which  to  draw  the  lines 
representing  the  unknown  forces.  Thus  the  polygon  is  completed 
by  drawing  from  C  and  A  lines  parallel  to  ck  and  ka  respectively, 
their  point  of  intersection  being  lettered  K.  Proceeding  to  the 
next  joint,  kc  and  cd  are  the  known  forces.  From  the  points  D 
and  K  lines  are  drawn  parallel  to  dj  and  jk  respectively,  inter- 
secting at  /,  thus  completing  that  polygon.  By  this  mode  of 
construction,  the  polygons  of  forces  for  the  various  joints  of  the 
truss  are  grouped  together  in  a  single  diagram  called  the  stress 
diagram.  When  the  last  joint  is  reached,  all  the  stresses  but  one 
will  be  known.  The  letters  representing  this  one  will  already 
be  in  the  stress  diagram,  and  the  line  joining  them  will  be  parallel 
to  the  corresponding  member  of  the  truss  if  the  construction  is 
accurate. 

The  following  points  should  be  noted  in  using  Bow's  notation: 

(1)  The  stress  in  any  member  of  the  truss  is  represented  by 
the  same  letters  as  the  member  itself. 

(2)  The  polygon  of  forces  for  any  joint  is  lettered  with  the 
letters  surrounding  that  joint,  the  direction  of  the  forces  being 
indicated  by  the  succession  of  letters  obtained  by  reading  the  letters 
surrounding  the  joint  in  right-handed  order. 

(3)  To  determine  whether  any  member  is  in  tension  or  com- 
pression,  read   the  letters   representing  that   member  in  right- 
handed  order  about  the  joint  at  either  end  of  the  member.     The 
same  order  of  succession  of  these  letters  in  the  stress  diagram 
indicates  the  direction  of  the  force  which  the  member  exerts  on 
the  joint  used.     If  this  direction  is  towards  the  joint,  the  stress  is 
compression;    if  away  from  it,  tension.     As  an  illustration,  let 
the  nature  of  the  stress  in  ck  be  determined.     Using  the  left  joint, 
the  letters  read  c-k.      In  the  stress  diagram  this  order  of  succes- 
sion of  these  letters  is  towards  the  left.     Referring  again  to  the 


ROOF-TRUSSES.  43 

truss,  the  direction,  towards  the  left,  is  seen  to  be  towards  the 
joint  used  as  a  centre,  this  indicating  that  the  member  ck  is  ex- 
erting force  on  this  joint  towards  it.  Hence  ck  is  in  compression- 

(4)  In  constructing  the  polygons  of  forces  it  is  to  be  noted 
that  all  the  sides  of  any  polygon  but  the  two  corresponding  to- 
the  two  unknown  forces  at  the  joint  will  be  already  represented 
in  the  stress  diagram,  and  the  points  from  which  to  draw  these 
two  sides  are  indicated  by  the  lettering,  as  previously  explained. 
The  point  of  intersection  of  these  two  sides  is  marked  with  the 
letter  common  to  the  corresponding  members  of  the  truss. 

(5)  Referring  to  Fig.  27,  any  portion  of  the  truss  (e  g.,  the 
shaded  portion)  is  a  body  in  equilibrium.     The  external  forces, 
acting  on  it  must  then  form  a  polygon  which  would  be  lettered 
with   the   letters   surrounding   it  (e.g.,  abcdjka).     It   should   be: 
observed  that  all  such  polygons  are  included  in  the  stress  diagram! 
(see  Fig.  28).     Hence  the  same  stress  diagram  would  be  obtained1 
whether  the  method  of  joints  or  method  of  sections  were  em- 
ployed in  constructing  the  various  force  polygons. 

In  using  the  method  illustrated,  the  truss  is  drawn  accurately 
to  scale,  the  loads  are  also  plotted  accurately  to  scale,  and  the 
stress  diagram  is  constructed  with  extreme  care.  The  magni-  ' 
tudes  of  the  stresses  are  gotten  by  scaling  off  the  proper  lines  of  the.- 
stress  diagram,  and  the  nature  of  the  stress  is  derived  from  the 
lettering,  as  previously  explained.  The  most  serious  cause  of 
inaccuracy,  especially  in  case  of  trusses  having  a  large  number  of 
members,  is  that  the  construction  of  each  polygon,  in  turn, -is 
based  upon  preceding  ones,  so  that  errors  accumulate.  To  guard 
against  this,  the  method  of  sections  should  be  employed  as  indi- 
cated in  Art.  37. 

The  student  is  recommended  to  study  trusses  with  the  view  of 
understanding  the  purpose  and  action  of  their  various  members 
independently  of  the  stress  diagrams.  In  Fig.  28,  for  example,  the 


44  GRAPHIC  STATICS. 

load  cd  tends  to  deflect  the  rafter,  this  deflection  being  prevented 
by  the  brace  kj,  which  would  therefore  be  in  compression,  kj, 
being  in  compression,  exerts  a  downward  thrust  on  the  lower 
chord  at  a.  The  lower  chord  is  prevented  from  deflecting  under 
this  thrust  by  ji;  ji  is  therefore  in  tension.  A  study  of  trusses 
in  such  a  manner  will  assist  the  student  to  a  better  understand- 
ing of  their  design. 

36.  Determination  of  Maximum  and  Minimum  Stresses.  In 
order  to  proportion  the  various  parts  of  a  truss,  it  is  necessary 
to  know  the  extreme  range  of  the  stresses  to  which  each  member 
will  be  subjected  under  the  various  combinations  of  loads  liable 
to  occur.  The  loads  to  take  into  account  will  vary  with  the  cir- 
cumstances. In  case  of  the  arches  supporting  the  train-shed  roof 
of  the  Philadelphia  and  Reading  Terminal  Station,  Philadelphia, 
the  stresses  were  determined  for:  (i)  dead  load;  (2)  dead  load  and 
snow  on  one  side;  (3)  dead  load  and  snow  on  both  sides;  (4) 
dead  and  wind  loads;  (5)  dead,  snow,  and  wind  loads.1  On  the 
other  hand,  the  train-shed  trusses  of  the  South  Terminal  Station, 
Boston,  were  designed  to  support  a  uniformly  distributed  vertical 
load  only.2  The  most  usual  combinations  are:  (i)  dead  load; 
(2)  dead  and  snow  loads;  (3)  dead  and  wind  loads. 

There  are  two  methods  of  procedure  in  this  connection:  (i) 
Construct  a  separate  stress  diagram  for  each  kind  of  loading,  and 
combine  the  stresses  thus  determined;  (2)  Construct  a  diagram 
for  each  combination  of  loads.  The  student  should  be  familiar 
with  both  methods;  the  latter,  however,  appears  to  be  generally 
preferable  for  the  following  reasons:  (i)  Under  certain  assump- 
tions the  first  method  does  not  give  correct  results;  e.g.,  under 
the  assumption  of  parallel  reactions,  if  the  reactions  for  dead 


See  Trans.  Am.  Soc.  C.  E.,  Aug.,  1895. 
1  Ibid.,  Dec.,  1899. 


ROOF- TRUSSES.  45- 

and  wind  loads,  considered  separately,  are  assumed  to  be  parallel,, 
the  resultant  reactions  found  by  combining  these  will  in  general, 
not  be  parallel.  Again,  in  dealing  with  friction  at  the  supports, 
the  two  methods  will  give  different  reactions,  and  consequently 
different  stresses.  (2)  In  dealing  with  counterbracing  (§  4), 
the  second  method  is  the  more  simple  of  the  two;  (3)  When 
the  second  method  is  employed,  the  maximum  and  minimum 
stresses  can  be  scaled  directly  from  the  diagrams. 

37.  Example.  Wooden  Truss  fixed  at  Both  Ends.  Figure 
i,  Plate  I,  represents  one  of  a  series  of  parallel  trusses,  spaced  16 
ft.  between  centres,  supporting  a  roof.  The  vertical  iron  tie-ro'ds 
divide  it  into  eight  panels  of  equal  width.  Span  =  80  ft.  Rise 
of  upper  chord  =16  ft.  The  roof  is  to  be  covered  with  tin  laid 
on  sheathing  i  in.  thick.  The  sheathing  is  supported  by  rafters 
2  in.  by  7  in.  section,  spaced  two  feet  between  centres.  The 
rafters  are  supported  by  purlins  8  in.  by  12  in.  section,  these 
being  supported  at  the  joints  of  the  upper  chord. 

Each  rafter  supports  an  area  of  the  roof  surface  2  feet  wide 
by  10.8  ft.  long,  and  is  proportioned  as  a  beam  to  support  the 
maximum  load,  including  its  own  weight,  which  is  liable  to  come 
upon  this  area.  Each  intermediate  purlin  supports  an  area 
10.8  ft.  wide  by  16  ft.  long,  and  is  also  proportioned  as  a  beam 
for  the  maximum  load  to  which  it  is  liable  to  be  subjected.  The 
student  should  verify  these  dimensions  of  rafters  and  purlins, 
using  a  working  stress  of  1000  Ibs.  per  sq.  in. 

CALCULATION  OF  LOADS. 

Wt.  of  tin  and  sheathing  per  sq.  ft.  of  roof  surface  =  3 £  Ibs. 
"    "   rafters  (30  Ibs.  per  cu.  ft.)      "     "         "         =  i  £    " 
"    "  purlins  (30  Ibs.  per  cu.  ft.)     "     "         "         =2     ' 

Total   =7    Ibs. 


46  GRAPHIC  STATICS. 

Estimated  weight  of  truss  =5760  Ibs.  This  weight  is  assumed 
to  be  supported  at  the  joints  of  the  upper  chord. 

(The  weight  of  the  truss  must  be  estimated  from  the  actual 
weights  of  trusses  of  similar  design  and  loading  already  built. 
Merriman  gives  the  formula  for  wooden  trusses,  W  =  %al(i  +  ^l), 
.and  for  iron  trusses  JF  =  fa/(i  +  Ty).  These  formulas  are 
•derived  from  a  table  of  weights  of  trusses  given  by  Ricker.  John- 
son gives,  for  iron  trusses,  the  formula  W=s\al2.  In  each 
case  W  =  weight  of  truss  in  pounds,  a  =  distance  between  trusses 
in  feet,  /  =  span  in  feet.  These  formulas  can,  at  the  best,  be  only 
rough  approximations.) 

The  snow  load  is  assumed  to  be  20  Ibs.  per  sq.  ft.  of  horizontal 
projection  of  roof  surface,  and  the  wind  pressure  40  Ibs.  per.  sq. 
ft.  on  a  vertical  surface. 

Normal  component  of  wind  pressure  (Hutton's  formula)  = 
20  Ibs.  per  sq.  ft.  of  roof  surface. 

LOADS  AT  INTERMEDIATE  JOINTS  or  UPPER  CHORD. 

Roof  covering,  etc  .........................   7X16X10.8=  1210  Ibs. 

Truss  ...................................  $X576o  =    720  " 


Total  dead  load  at  each  intermediate  joint  =  1930  Ibs. 

Snow  "  =20X16X10     =3200  " 

Wind  "     "      "  "  "  =20X16X10.8  =  3460  " 


Each  end  joint  supports  one-half  the  load  carried  by  an  inter- 
mediate joint. 

The  loadings  considered  in  this  example  are:  (i)  dead  load; 
(2)  dead  and  snow  loads;  (3)  dead  and  wind  loads.  The  stu- 
dent should  trace  out  the  construction  of  the  diagrams,  noting 
the  steps  taken  to  secure  accuracy  and  to  check  the  work. 

Dead-load  Diagram  (Fig.  i  B).  The  reaction  of  each  sup- 
port is  7720  Ibs.  The  total  load  is  plotted  and  subdivided  to 


ROOF-TRUSSES.  47 

obtain  the  joint  loads.  Since  the  truss  is  symmetrical  and 
symmetrically  loaded,  the  stresses  in  corresponding  members 
on  the  two  sides  will  be  equal;  the  diagram  is  therefore  con- 
structed for  one  side  only.  In  order  to  avoid  a  long  succession  of 
joints  and  to  check  the  work,  the  stresses  in  er,  rq,  and  qk  are 
determined  at  the  outset  by  the  method  of  sections  (Art.  32)  as 
follows : 

Calculate  the   stress  in   one  of  these  members,   as  qk,  by 
moments.     Taking  the  moment  axis  at  the  intersection  of  the 
other  two  members,  we  have 
7720-30—  965-30—  1930(20+  io)  =  i2QK.       .'.  QK=  12,062  Ibs. 

The  moment  of  the  known  forces  being  right-handed,  the 
moment  of  QK,  to  balance,  must  be  left-handed,  hence  QK  is 
in  tension  since  the  force  acts  away  from  the  portion  of  the  truss 
under  consideration.  Next  plot  this  computed  stress  in  the 
diagram,  laying  off  from  K  the  length  QK  in  such  direction  that 
it  will  represent  tension.  The  stresses  in  er  and  rq  are  next 
determined  by  completing  'the  polygon  QKABCDERQ.  The 
remaining  stresses  to  the  left  of  the  section  are  determined  by  the 
method  of  joints  as  follows: 

At  joint  i  determine  BL  and  LK;  at  joint  2,  LM  and  MK; 
at  joint  3,  CN  and  NM;  at  joint  4,  DP  and  PQ;  at  joint  5,  PO 
and  OK.  If,  now,  the  closing  line  NO  is  parallel  to  the  member 
no,  the  work  is  checked.  By  this  order  of  procedure  a  long  suc- 
cession of  force  polygons  dependent  on  each  other  has  been 
avoided,  thus  preventing  inaccuracies  from  accumulating,  and 
the  work  has  been  completely  checked.  Trusses  should  be 
divided  in  this  manner  by  a  sufficient  number  of  sections  so  that 
each  division  will  contain  not  more  than  ten  or  twelve  joints. 
It  should  be  noted  that  the  mere  closing  of  the  force  polygons, 
without  the  introduction  into  the  stress  diagram  of  stresses  com- 
puted independently  of  the  diagrams,  does  not  completely  check 


48  GRAPHIC  STATICS. 

the  work,  since  errors  made  in  constructing  the  truss  diagram 
and  plotting  th:  loads  would  not  interfere  with  the  diagrams 
closing. 

Dead  and  Snow  Loads.  Since,  in  this  example,  the  snow  load 
is  distributed  in  the  same  manner  as  the  dead  load,  the  stresses 

can  be  found  by  multiplying  the  dead-load  stresses  by  -^  -  ~  —  • 

Dead  and  Wind  Loads  (Fig.  i  A).  The  wind  is  taken  to 
come  from  the  right.  Attention  is  directed  to  the  following 
points: 

(1)  The  resultant  dead  and  wind  loads  are  first  plotted,  as 
shown  by  the  dotted  lines,  the  joint  loads  being  found  by  suitable 
subdivision. 

(2)  The  resultant  load  intersects  the  lower  chord  at  z,  the 
reactions  JK  and  KA  (assumed  to  be  parallel)  being  found  by 
dividing  AJ  into  parts  inversely  proportional  to  the  segments 
of  the  lower  chord. 

(3)  The  stresses  are  determined  in  a  manner  similar  to  that 
explained  for  the  dead  load.     In  calculating  the  stress  in  qk  it  is 
convenient  to  know  the  H  and   V  components  of  the  external 
forces. 

H  component  of  wind  load  =  -  ^  •  13840  =   5  140  Ibs. 

40 
V  component  of  wind  load=  -  g  •  13840=  12850  Ibs. 

Taking  moments  about  the  right-hand  support, 

80  Vl=  1  5440  -40+  1  2850  -20+  5  140  -8.    .'.  F1  =  ii4461bs. 


Hl  =  -   -^  —  1  1446  =  2080  Ibs. 

15440+  12850 


ROOF-TRUSSES. 


49 


QK  = 


1 1446  •  30—  2O8O  -12  —  965  •  30—  1930(20+  IO) 


12 


19300  Ibs.  (tension). 


Maximum  and  Minimum  Stresses.  The  maximum  and 
minimum  stresses  for  the  different  members  of  the  truss  were 
determined  as  follows:  Tabulate  the  stresses  in  all  the  members, 
for  each  of  the  three  combinations  of  loads.  Note  that  the 
extreme  range  of  stresses  will  be  the  same,  in  case  of  correspond- 
ing members  on  the  two  sides  of  the  truss.  Thus,  considering 
the  members  op  and  uv,  we  find  the  stresses  (in  pounds)  re- 
corded in  the  following  table: 


Dead. 

Dead  and 
Snow. 

Dead  and 
Wind. 

uv 

+  3100 
+  3100 

+  8200 
+  8200 

+  3100 
+  9100 

Of  these  stresses,  the  extremes  are  9100  Ibs.  and  3100  Ibs- 
(see  Fig.  i  C).  Each  of  these  two  members  must  then  be  designed 
to  support  a  compression  stress  of  9100  Ibs.,  the  minimum  stress, 
3100  Ibs.,  being  also  used  if  the  effect  of  repetition  of  stress  is 
considered  in  such  design. 

The  maximum  and  minimum  stresses  in  all  the  members 
are  recorded  in  Fig.  i  C.  These  should  be  verified  by  the 
student. 

If  the  truss  is  unsymmetrical,  supported  differently  at  the 
two  ends,  etc.,  the  above  procedure  must  evidently  be  modified. 

38.  General  Remarks,  i.  When  the  loading  is  complicated, 
special  precautions  should  be  taken  to  plot  the  joint  loads  in 
such  a  manner  as  to  avoid  any  accumulation  of  errors.  This  may 
be  accomplished  as  indicated  in  Art.  37,  otherwise  as  follows  (see 


5° 


GRAPHIC  STATICS. 


Fig.  29) :  Having  resolved  the  loads  into  H  and  V  components,  plot 
D<     AD'  =  ZH     and     D'D  =  2V,     then 
AD  is  the  resultant  load.     To  plot 
/   the  joint  loads,  lay  off  on  AD'  the 
H  component  joint  loads  AB',B'Cf, 
etc.,  in  order,  and    on  D'D  the   V 
'  component  joint  loads  D'B" ',  B"C", 
etc.,  in  the  same  order.     Combining 
these  we  have  AB,  BC,  etc.,  for   the 
resultant  joint  loads. 

2.  The  precaution,  illustrated  in 
Art.  37,  of  computing  the  stress  in 
one  or  more  of  the  members  by  the 


FIG.  29. 


method  of  sections,  and  plotting  the  results  in  the  stress  diagram, 
should  always  be  employed  in  large  or  important  work. 

3.  In  constructing  stress  diagrams  it  is  sometimes  desirable 
to  obtain  the  directions  of  the  members  more  accurately  than  can 
be  done  by  using  the  truss  diagram  unless  the  latter  is  constructed 
on  an  inconveniently  large  scale.      For  this  purpose  coordinate 
axes  can  be  used,  the  slopes  of  the  various  members  being  plotted 
on  them  as  indicated  for  st  and  uv  (Fig.  i  D).     A  good  general 
rule  to  follow  is  to  avoid  obtaining  the  direction  of  a  line  of  the 
stress  diagram  from  a  shorter  line  of  the  truss. 

4.  Nothing  is  gained  by  constructing  the  stress  diagram  on 
a  very  large  scale.     The  lines  of  such  diagrams  need  not  be  in 
error  more  than  two  or  three  hundredths  of  an  inch.     With  a 
scale  of  10,000  Ibs.  to  the  inch  this  would  be  an  error  of  200  to 
300  Ibs.  in  the  stress.     Such  an  error  is  of  no  importance  in 
work  requiring  so  small  a  scale. 

5.  The  diagrams  should  be  constructed  with  such  care  and 
in  such  a  manner  as  to  leave  no  question  as  to  their  accuracy  at 
any  stage  of  the  construction.     The  wooden  edges  of  the  ordinary 


ROOF-  TRUSSES.  5 1 

drawing-board  and  T  square  are  unreliable.  A  steel  straight- 
edge with  lead  weights  to  hold  it  in  position  is  preferable.  The 
edges  of  the  triangles  must  be  straight  and  the  90°  angle  true. 
The  usual  hard -rubber  and  celluloid  triangles  are  not  sufficiently 
accurate  in  these  respects  and  should  be  tested  before  using. 
The  best  quality  of  dividers  and  compasses,  a  metal  scale  grad- 
ated to  hundredths  of  an  inch,  and  a  hard  pencil  kept  sharpened 
to  a  fine  chisel  edge,  complete  the  list  of  necessary  instruments. 
All  intersections  to  be  preserved  are  located  by  a  fine  prick-point 
enclosed  in  a  circle. 

39.    Problems,     i.  The  truss,  Fig.   30,  fixed  at  both  ends, 
is  one  of  a  system  of  parallel  trusses,  spaced  10  ft.  apart,  support- 


ing a  roof.  The  dead  load  is  20  Ibs.  per  sq.  ft.  of  roof  surface. 
The  wind  pressure  is  assumed  to  be  40  Ibs.  per  sq.  ft.  on  a  vertical 
surface.  Determine  the  stresses  in  all  the  members  due  to  the 
combined  dead  and  wind  loads. 

Directions,  (i)  Assume  the  H  component  reactions  to  be 
equal.  (2)  Determine  the  normal  wind  loads  on  the  two  inclined 
roof  surfaces  by  Hutton's  or  Duchemin's  formula  (see  table,  p.  32). 
(3)  Plot  these  two  wind  loads  together  with  the  resultant  dead 
load  and  determine  the  reactions  of  supports  by  drawing  a  funicu- 
lar polygon  for  these  (three)  resultant  loads  (see  Arts.  15  and 


5  2  GRAPHIC  STATICS. 

29).  (4)  Plot  the  joint  loads  in  order  (see  Art.  37)  and  draw  the 
stress  diagram,  checking  on  the  middle  portion  of  the  lower 
chord. 

PROBLEM  2.     The  truss,  Fig.  31,  is  loaded  on  the  upper  chord 
with  a  uniformly  distributed  dead  load  of  75,000  Ibs.     In  addi- 


tion,  each  intermediate  joint  of  the  lower  chord  is  loaded  with 
2000  Ibs.  The  lower  chord  is  a  circular  arc.  The  vertical  web 
members  are  equidistant.  Determine  the  stresses  in  all  the  mem- 
bers. 

Directions,  (i)  In  plotting  the  truss,  locate  the  joints  of 
the  lower  chord  by  ordinates.  (2)  Plot  the  external  forces  in 
right-handed  order  as  usual,  namely,  loads  on  upper  chord — 
right-hand  reaction — loads  on  lower  chord — left-hand  reaction. 
(3)  Calculate  stress  in  member  a  (or  a')  by  moments,  plotting 
it  in  the  stress  diagram.  (4)  Construct  the  stress  diagram  for 
either  half-truss,  checking  on  b  (or  &'). 

PROBLEM  3.  Determine  the  maximum  and  minimum  stresses 
in  the  truss,  Fig.  31,  providing  for  a  wind  pressure  of  40  Ibs. 
per  sq.  ft.  on  a  vertical  surface  as  well  as  for  the  dead  loads  of 
Prob.  2.  Assume  that  one  end  of  the  truss  rests  on  rollers,  the 
reaction  at  that  end  being  vertical.  (Note.  Three  combina- 
tions of  loads  must  be  dealt  with,  namely:  dead  load  alone; 


ROOF-TRUSSES. 


53 


dead  load  with  wind  on  roller  side;    dead  load  with  wind  on 
fixed  side.) 

40.  Cantilever  Truss.     Fig.  32,  which  represents  one  of  the 
train-shed  trusses  of  the  South  Terminal  Station,  Boston,  con- 


FIG.  32. 

sists  of  a  central  truss,  AB,  supported  at  A  and  B  on  the  ends  of 
the  cantilevers.  AC,  acting  as  a  link,  provides  for  expansion. 
The  dotted  chord  members  are  furnished  with  sliding  joints  and 
thus  do  not  resist  tension  or  compression.  The  stresses  can  be 
determined  in  the  usual  way,  first  finding  the  reactions  at  A 
and  B  due  to  the  loads  on  the  central  truss.1 


41.  Trusses  having  only  Two   Forces  to  determine  at  Each 
Joint.    In  the  preceding  examples  only  two  unknown  stresses 
were  encountered  at  each  joint,  so  that  the  polygons  of  forces  for 
the  successive  joints  of  the  truss  could  be  constructed  at  once. 
Other  cases  of  the  same  nature  are  given  in  Fig.  33  A,  B,  C,  D, 
E,  F,  G,  and  H.     The  student  should  verify  this  statement,  in- 
dicating the  order  of  succession  of  joints  to 'use  in  constructing 
the  stress  diagrams. 

42.  Fink,  or  French,  Roof-truss.     This  is  a  type  of  iron  truss 
in  common  use  for  shops  and  similar    buildings.      Fig.  33  L 
represents  its  most  usual  form.     In  applying  the  method  of  joints 

1  See  Trans.  Am.  Soc.  C.  E.,  Dec.,  1899. 


GRAPHIC  STATICS. 


FIG.  33 


ROOF-TRUSSES.  55 

to  this  truss  a  difficulty  is  encountered  to  which  attention  is 
directed.  Beginning  at  the  left-hand  support,  the  force  poly- 
gons can  be  constructed  in  the  usual  manner  until  the  joint  S 
is  reached,  where  three  unknown  forces  are  encountered.  There 
are  various  graphical  devices  for  overcoming  the  difficulty;  the 
most  satisfactory  solution,  however,  is  to  calculate  the  stress  in 
ST  by  moments  (method  of  sections)  and  insert  this  calculated 
stress  in  the  diagram.  The  .stresses  in  the  two  remaining  mem- 
bers at  the  joint  5  can  then  be  found  by  completing  the  force 
polygon.  If  the  method  of  procedure  of  Art.  37  is  followed 
no  difficulty  arises.  Other  trusses  of  the  same  nature  are 
shown  in  Fig.  33  I,  J,  K,  and  M. 


§  4.     Counterbracing  and  Double  Systems  of  Bracing. 

43.  Definitions.  It  will  be  noticed  that  the  members  of  a 
properly  designed  truss  form  in  general  a  system  of  triangles. 
The  triangle  is  the  elementary  truss.  Under  the  action  of  forces 
lying  in  its  plane  and  acting  at  its  vertices,  it  cannot  be  distorted 
without  changing  the  lengths  of  one  or  more  of  its  sides,  and 
such  changes  are  opposed  by  the  resistance  to  extension  and 
compression  of  the  sides.  A  polygonal  frame  of  more  than  three 
sides,  assumed  free  to  turn  at  the  joints,  can  be  distorted  with- 
out altering  the  lengths  of  any  of  the  sides. 

The  quadrilateral  frame  (Fig.  34),  acted  on  by  the  force  Ff 
would    be    distorted    as    shown,   the   distance^ 
AB  becoming  shorter  and  CD  longer,  change 
in  length  of  the  diagonals  necessarily  accom- 
panying the  distortion  of  the  frame.     A  diag- 
onal member  capable  of  resisting  both  extension 
and    compression  would,  then,  prevent  distor- 
tion.    A  member  joining  C  and  D,  capable  of  resisting  tension 


$6  GRAPHIC  STATICS. 

alone,  would  prevent  distortion  in  the  direction  shown  in  Fig.  34, 
but  not  in  the  opposite  direction.  Two  tension  diagonals,  however, 
would  evidently  make  the  frame  stable.  Thus  a  quadrilateral 
frame  may  be  made  capable  of  resisting  any  forces  acting  at  the 
angles,  tending  to  distort  it,  by  the  introduction  of  a  single  diagonal 
.  member  capable  of  resisting  both  tension  and  compression,  or 
of  two  diagonals,  both  capable  of  resisting  tension  alone  or  com- 
pression alone.  In  the  latter  case  it  is  evident  that  only  one  of 
the  diagonals  would  be  stressed  at  a  time.  Considering  the  quad- 
rilateral of  Fig.  34  to  represent  one  panel  of  a  truss,  the  diagonal 
which  is  stressed  under  the  action  of  the  dead  load  may  be  called 
the  main  brace,  the  other  the  counterbrace  or  counter.  The  counter 
may  be  stressed  under  the  action  of  the  wind,  snow  on  one  side 
of  the  roof,  or  other  non-symmetrical  temporary  load. 

44.  Notation.    The  system  of  notation  of  Fig.   35   is   con- 
venient to  use  for  trusses  with  counterbracing.     One  diagonal 
of  each  panel  is  drawn  dotted.     The  diagonals  drawn  in  full 
lines  are  designated  by  the  letters  gh  and  ij,  the  same  letters 
accented  being  used   for  the  dotted  diagonals.     To   illustrate, 
if  the  diagonals  under  stress  were  g'h'  and  ij,  the  verticals  would 
be  represented  by  ]hf,  gfi,  and  jk;    the  members  of  the  upper 
chord  would  be  bgf  and  a,  and  of  the  lower  chord  h'e  and  je, 
the  diagonals  not  stressed  being  considered  omitted  from  the 
diagram. 

45.  Determination  of  Stresses.     Diagrams  Drawn   for  Com- 
bined Loads.     CASE  i.     Parallel  Chords.     Having  plotted  the 
polygon    of    external    forces,    we   must    first    determine    which 
diagonal  in  each  panel  is  stressed  under  the    given    system    of 
loads.     In  the   case   of  parallel  chords  this   may  be    done    as 
follows  (see  Fig.  35): 


ROOf  TRUSSES. 


57 


1000 


Using  the  method  of  sections,  apply  the  condition  of  equilibrium, 
-TF  =  o,  to  the  forces  acting  on 
the  shaded  portion  of  the  truss. 
The  V  component  stresses  in  the 
chord  members,  bg  and  he,  are 
each  zero,  hence  the  V  compo- 
nent stress  in  the  diagonal  (gh  or 
gfhf)  under  stress  must  be  equal 
and  opposite  in  direction  to  the 
V  component  of  the  resultant  ex- 
ternal force.  Now  it  will  be  seen 
that  this  resultant  force  is  repre- 
sented in  the  stress  diagram  by 
the  letters  which  lie  below  and  above  the  panel  in  question,  read 
in  right-handed  order,  i.e.,  EB.  This  order  of  letters  (see  stress 
diagram)  is  downward,  indicating  that  the  resultant  external 
force  acts  downward,  hence  the  V  component  diagonal  stress 
must  act  upwards.  Keeping  in  mind  the  fact  that  if  a  member 
is  in  tension  the  force  it  exerts  upon  the  shaded  portion  of  the 
truss  acts  away  from  that  portion  and  conversely,  we  see  that  if  the 
diagonals  are  tension  members  gh  is  the  one  under  stress,  while 
if  they  are  compression  members  g'h'  is  stressed.  Thus  by 
observing  whether  the  resultant  external  force  lying  to  the  left 
of  any  panel  acts  up  or  down  as  indicated  in  the  external  force 
polygon,  and  knowing  whether  the  diagonals  are  tension  or  com- 
pression members,  we  can  determine  at  a  glance  which  diagonal 
is  under  stress.  The  use  of  the  condition  2T  =  o  in  determining 
stresses  is  called  the  method  of  shears. 

In  Fig.  35  the  diagonals  are  assumed  to  be  tension  members, 
those  drawn  in  full  lines  being  consequently  under  stress.  The 
dotted  diagonals  are  therefore  omitted  and  the  stress  diagram 
constructed  in  the  usual  manner. 


GRAPHIC  STATICS. 


CASE  2.  Chords  not  Parallel,  In  this  case  the  condition  of 
equilibrium,  JTM  =  o,  can  be  used,  as  indicated  in  the  following 
example,  to  determine  which  diagonals  are  stressed.  Having 
plotted  the  loads  (see  Fig.  36),  -the  reactions  DE  and  EA 


\  --v'" 
=y 


FIG.  36. 

(assumed  to  be  parallel)  are  determined  by  drawing  a  funicular 
polygon  for  the  joint  loads.  Now,  consider  the  panel  eb  and  the 
forces  acting  on  the  portion  of  the  truss  to  the  left  of  the  section 


ROOF-TRUSSES.  59 

mn.  The  two  chord  members  intersect  at  O'.  Taking  this 
point  as  moment  axis,  the  moment  of  the  stresses  in  the  members 
cut  by  mn  must  balance  the  moment  of  the  external  forces  acting 
on  the  left  portion  of  the  truss.  The  moment  of  each  chord 
stress  being  zero,  the  moment  of  the  stress  in  the  diagonal 
gh  or  g'h'  must  balance  the  moment  of  the  resultant  external 
'force  EB.  This  resultant  force  must  act  through  the  intersection 
of  the  strings  e  and  b  (see  Arts.  8  and  9).  The  intersection 
of  these  strings  falls  here  outside  the  limits  of  the  drawing, 
but  it  is  evident  from  the  direction  of  EB  and  the  general 
location  of  the  intersection  of  the  two  strings  e  and  b  that  the 
moment  of  EB  about  O'  would  be  right-handed;  hence  the  mo- 
ment of  the  diagonal  stress  must  be  left-handed,  i.e.,  the  diagonal 
gh  would  be  stressed  if  the  diagonals  were  tension  members;  and 
g'h'  if  they  were  compression  members.  Dealing  with  the  panel 
ec  in  the  same  manner,  we  find  that  the  moment  of  the  resultant 
external  force  EC  =  R,  about  the  axis  O,  is  left-handed,  hence 
the  moment  of  the  diagonal  stress  must  be  right-handed;  i.e., 
ij  is  stressed  if  the  diagonals  are  tension  members  and  con- 
versely. 

In  Fig.  36,  the  diagonals  are  assumed  to  be  tension  members, 
those  drawn  in  full  lines  being  stressed.  The  stress  diagram  can 
now  be  constructed  in  the  usual  manner. 

If  it  were  necessary  to  locate  the  resultant  force  EB,  it  could 
be  done  by  Art.  9  (Special  Case)  as  follows:  Using  A  for  pole, 
the  strings  e',  p',  and  V  of  the  special  polygon  for  the  forces  EP 
and  PB  are  drawn,  the  intersection  of  ef  and  bf  (not  shown)  being 
one  point  in  the  line  of  action  of  EB. 

It  should  be  remarked  that  the  diagonals  under  stress  may 
be  determined  by  trial  during  the  construction  of  the  stress 
diagram  as  follows:  Omit  either  diagonal  of  a  panel  at  random 
and  draw  the  force  polygons,  determining  from  the  order  of  the 


60  GRAPHIC  STATICS. 

letters  the  kind  of  stress  in  the  diagonal  used.  If  this  agrees 
with  the  stress  for  which  the  diagonals  were  designed,  the  diagram 
is  correct,  otherwise  the  other  diagonal  is  stressed  and  the  force 
polygons  involved  must  be  re-drawn.  This  method  is  unsatis- 
factory compared  with  those  previously  described. 

46.  Example.  The  lower  chord  of  the  truss,  Fig.  2,  Plate  I, 
is  a  circular  arc  of  185  ft.  radius.  The  verticals  are  equidistant 
and  the  diagonals  are  tension  members.  Both  ends  are  fixed 
and  the  reactions  are  assumed  to  be  parallel.  It  is  required  to 
determine  the  stresses  in  all  the  members  for  the  following  system 
of  loads,  viz.:  i.  A  uniformly  distributed  dead  load  of  75,000  Ibs., 
the  loading  (18,750  Ibs.)  on  the  two  central  panels  being  divided 
between  the  monitor  and  main  roofs  as  follows:  5000  Ibs.  on 
the  monitor  roof  and  13,750  Ibs.  supported  directly  by  the  main 
roof.  2.  A  normal  wind  pressure  on  the  left  side,  distributed  as 
follows:  17,000  Ibs.  on  vertical  surface  of  main  roof;  33,000  Ibs. 
on  inclined  surface  of  main  roof;  9600  Ibs.  on  vertical  surface  of 
monitor  roof,  and  11,000  Ibs.  on  inclined  surface  of  monitor  roof. 

1.  Stresses   in   Monitor   Roof.     The   joint    loads    and    stress 
diagram  are  given  in  Fig.  2  B.     The  diagonals  are  assumed  to 
be  tension  members,  }e  being  evidently  the  one  under  stress; 
gh  is  therefore  omitted.     The  diagram  is  constructed  beginning 
at  the  joint  ab.     The  stresses  in  the  members  ah,  hg,  g},  je,  and 
ed  evidently  act  as  loads  on  the  main  truss  and  are  included  in 
the  table  of  joint  loads  given  below.     The  computed  stress  in 
g}  is  50  Ibs.  tension. 

2.  Joint  Loads  and  Reactions  for  Main   Truss.     The  joint 
loads  for  the  main  truss,  resolved  into  H  and  V  components  for 
convenience  in  plotting  and  computing,  are  given  in  the  following 
table: 


ROOF-TRUSSES. 
TABLE  or  JOINT  LOADS  FOR  MAIN  TRUSS. 


6r 


Joints. 

bbi 

bic 

cd 

de 

*f 

fg 

fh 

hi 

v 

j* 

H  loads 

8500 

458? 

4^83 

7092 

0383 

V  loads  

o 

9687 

19375 

J9375 

22625 

6121 

14379 

9375 

9375 

4687 

Having  plotted  the  joint  loads  (see  Fig.  2  A),  the  reactions 
KA  and  ^45  are  determined  by  drawing  a  funicular  polygon  for 
the  resultant  joint  loads  as  shown.  (In  drawing  such  a  polygon: 
begin  with  the  middle  string  /  or  g.) 

3.  Determination    of    Diagonals    Stressed.     The    method    of 
moments  (Art.  45,  Case  2)  is  used  to  determine  which  diagonal 
in  each  panel  is  stressed.     E.g.,  in  panel  c,  the  resultant  ex- 
ternal force  for  the  portion  of  the  truss  to  the  left  of  the  section 
mn  is  R  =  AC,  acting  through  the  point  of  intersection  of  the 
strings  a  and  c.     The  point  of  intersection  of  the  two  chord 
members  is  not  shown  but  evidently  lies  to  the  left  of  R,  so  that 
the  moment  of  R  about  this  point  is  left-handed.     The  moment; 
of  the  diagonal  stress  must   then   be   right-handed.     Therefore, 
as    the    diagonals    are    tension  members,  the  one  in  full  line  is. 
stressed.     Investigating   each   panel  in  a  similar  manner,  it  is 
found   that  the  diagonals  stressed  are  zi,  yx,  w'i/ ,  u't',  s'r*,  qp, 
on,  and  ml.     (See  notation  of  Art.  44.) 

4.  Calculation  by  Method  of  Sections.     The  stress  in  gs?  is 
now  calculated  by  moments  as  follows:    Having  previously  de- 
termined that  the  dotted  diagonal  s'r1  is  the  one  stressed,  we 
take  as  moment  axis  the  intersection  of  this  diagonal  and  the 
lower  chord  member,  i.e.,  the  joint  rp.     The  computed  stress 
is  77,330  Ibs.  compression. 

It  should  be  observed  that  it  is  necessary  to  know  which  diago- 
nal is  stressed  in  order  to  make  this  calculation. 

5.  Construction    of   Stress    Diagram.     The    stress    in    gsf    is 
plotted  in  the  diagram  and  the  diagram  is  completed  in  the  usual 


62 


GRAPHIC  STATICS. 


manner,  working  in  both  directions  from  centre  and  ends  and 
checking  midway.  The  results  are  given  in  the  table,  Plate  I. 

It  should  be  noted  that  no  counterbrace  is  needed  in  those 
panels  where  the  same  diagonal  is  stressed  under  all  circumstances. 

PROBLEM.  In  the  truss,  Fig.  2,  Plate  I,  determine  which  diago- 
nals are  stressed  under  the  action  of  the  dead  load  alone. 


47.  Determination  of  Stresses  in  Trusses  with  Counterbracing. 
SECOND    METHOD.     When   separate    diagrams   are    constructed 


FIG.  37. 


for  the  different  kinds  of  loading  (see  Art.  36),  the  determination 
of  the   maximum  stresses,  where  counterbracing  is  involved,  is 


ROOF-TRUSSES.  63 

i 

somewhat  complicated.  A  method  of  procedure  is  illustrated 
in  the  example,  Fig.  37.  The  notation  of  Art.  44  is  employed. 
The  diagonals  are  assumed  to  be  tension  members.  The  load- 
ings are:  i.  Dead  load;  2.  Wind  load.  Both  ends  of  the  truss 
are  fixed,  the  H  reactions  being  assumed  equal.  Fig.  37  A  is 
the  diagram  for  wind  load  on  the  left  side,  Fig.  37  B  is  the  dead- 
load  diagram. 

1.  In  constructing  the  stress  diagrams,  one  set  of  diagonals, 
e.g.,  those  dotted,  is  omitted,  and  the  force  polygons  are  drawn 
in  the  usual  manner.     The  construction  is  then  repeated  in  the 
same  figure,  using  the  diagonals  previously  omitted  (the  addi- 
tional lines  are  dotted).     The  diagrams  thus  contain  the  stresses 
in  all  the  members  of  the  truss,   whichever  diagonal  of  each 
panel  is  assumed  to  be  stressed.     There  will  be  no  confusion 
if  the  mode  of  lettering  of  Art.  44  is  employed. 

2.  The  next   step   is  to  determine  which  diagonal  of  each 
•panel  is  stressed  under  each  combination  of  loads,  i.e.,  i.  Dead 
load  alone;  2.  Dead  load  and  wind  on  one  side. 

Dead  Load  Alone.  From  Fig.  37  B  we  find  that  the  diagonals 
in  tension  are  //  (full)  and  k'V  (dotted).  The  stresses  in  the  mem- 
bers are  therefore  lettered  in  Fig.  37  B  as  follows: 

BH,  CI,  DK',  EM,  HG,  JG,  L'G,  MG,  HJ,  //,  IKf,  K'L',  L'M. 

Dead  Load  and  Wind  on  Left  Side.  We  first  determine  which 
diagonal  of  each  panel  is  stressed  by  combining  the  stresses  of 
Figs.  37  A  and  37  B.  We  thus  find  that  the  diagonals  intension 
are  i'j'  (dotted)  and  WV  (dotted).  The  stresses  for  this  com- 
bination of  loads  are  therefore  lettered  in  the  two  diagrams  as 
follows : 

BH,  CF,  DK',  EM,  HG,  J'G,  L'G,  MG,  HI',  I'J',  J'K', 
K'V,  L'M. 


64  GRAPHIC  STATICS. 

The  resultant  stresses  are  therefore  found  by  combining  the 
stresses  lettered  as  above,  in  the  two  diagrams. 

Dead  Load  and  Wind  on  Right  Side.  In  this  example,  this 
combination  of  loads  need  not  be  dealt  with  as  it  is  obvious  that 
the  maximum  stresses  in  corresponding  members  on  the  two  sides 
of  the  truss  will  be  equal. 

3.  Having  determined  the  stresses  due  to  the  different  com- 
binations of  loads,  the  maximum  and  minimum  stresses  can  be 
selected  as  explained  in  Art.  37. 

& 

48.  Trusses  having  a  Double  System  of  Web  Members. 
Such  a  truss  may  be  treated  as  a  combination  of  two  trusses  hav- 
ing common  chords,  but  distinct  systems  of  web  members.  The 
girder  of  Fig.  38  A,  for  example,  can  be  resolved  into  those  shown 
in  Figs.  38  B  and  38  C,  the  stresses  in  the  members  of  these  com- 
ponent trusses  being  found  in  the  usual  manner.  The  actual 

stress  in  any  web  member  of 
the  original  truss  is  given 
directly  by  the  diagrams  while 
the  stress  in  any  chord  seg- 
ment, as  ab,  is  evidently  equal 
to  the  algebraic  sum  of  the 
stresses  found  for  cd  (Fig.  38 
B)  and  ef  (Fig.  38  C).  When 
the  given  truss  can  be  resolved 
in  more  than  one  way,  or  when 
the  distribution  of  loads  be- 
tween the  component  trusses 
is  uncertain,  the  problem  is 
indeterminate. 

Fig.  39  is  a  crescent  roof-truss  with  two  systems  of  web  mem- 
bers, as  shown  by  the  full  and  dotted  lines  respectively.  By 


ROOF- TRUSSES.  65, 

tracing  out  the  force  polygons  for  the  different  joints  it  will  be 
seen  that  the  stress  diagram  can  be  drawn  at  once  for  the  com- 
plete truss.  A  suitable  system  of  lettering  the  interior  of  the 
truss  is  given,  the  intersections  of  the  diagonals  being  treated 
as  joints.  If  the  joint  i  were  made  to  coincide  with  2,  the  por- 
tion of  its  load  supported  by  each  component  truss  would  be 


FIG.  39. 

uncertain  and  the  problem  in  this  particular  would  be  indeter- 
minate. 

49.  Double  Diagonal  Bracing.  In  the  case  of  counterbraced 
panels  (see  Fig.  33  N,  O,  P,  Q),  the  two  diagonals  are  assumed 
not  to  be  in  action  for  the  same  loading;  both  being  designed  for 
the  same  kind  of  stress.  When  the  diagonals  are  designed  to 
act  simultaneously  (one  being  in  tension  while  the  other  is  in 
compression),  the  truss  can  be  resolved  into  two  trusses  having 
common  chords  and  verticals  but  distinct  systems  of  diagonals. 
Each  joint  would  belong  to  both  component  trusses,  and  the 
division  of  its  load  between  the  two  trusses  would  be  uncertain. 
One  way  of  dealing  with  this  case  is  to  assume  that  the  load  at 
each  joint  is  divided  equally  between  the  two  trusses.  The 
stresses  in  the  verticals  and  chords  would  be  the  algebraic  sum 
of  those  found  for  the  separate  trusses.  The  results  of  this 
method  are  evidently  liable  to  err  in  the  wrong  direction. 


66 


GRAPHIC  STATICS. 


5 .     Three-  hinged  A  rch. 


50.  Definition.     A  three-hinged  arch  consists  of  two  arched 
ribs  hinged  at  the  crown  and  abutments  (Fig.  40).     The  outward 
thrust  at  the  ends  may  be  resisted  by  the  abutments  or  by  a  tie- 
rod  joining  the  ends.     The  ribs  may  be  braced  or  solid. 

51.  Determination   of   Reactions  of  Hinges.     The   reactions 
are  assumed  to  act  through  the  centres  of  the  hinges.     The  re- 
actions of  the  hinge  at  the  crown  on  the  two  half -ribs  must  be 
evidently  equal  and  opposite.     Let  ab  (Fig.  40)  represent  the 
line  of  action  of  the  resultant  load  supported  by  the  left  half-rib, 
the  right  half  being  assumed  to  be  unloaded.     The  reactions 
at  Of  and  O"  must  be  equal  and  act  along  the  line  O"O' .     The 
three  forces  acting  on  the  left  half  must,  for  equilibrium,  inter- 
sect at  a  common  point.     This  condition  determines  On  to  be 
the  direction  of  the  reaction  at  O. 


The  magnitudes  of  these  reactions  can  now  be  found  by  con- 
structing the  triangle  of  forces  ABD'.     BDr  is  the  magnitude 


ROOF-TRUSSES.  67 

of  the  reaction  at  O",  and  of  the  two  equal  and  opposite  re- 
actions at  O'.  D'A  is  the  magnitude  of  the  reaction  at  O. 

The  directions  and  magnitudes  of  the  reactions  at  O,  O', 
and  O",  for  any  resultant  load  be  on  the  right  half-rib,  are  found 
in  a  similar  manner,  the  triangle  of  forces  being  BCD".  When 
both  sides  are  loaded,  the  reaction  at  either  hinge  is  evidently 
the  resultant  of  the  reactions  due  to  the  loads  taken  separately. 
Combining  the  two  separate  reactions  for  each  of  the  hinges 
by  the  triangle  of  forces,  we  find  the  resultant  reactions  for  the 
right  half-rib  to  be  CD  and  DB  at  O"  and  O'  respectively;  and 
for  the  left  half- rib,  BD  and  DA  at  O'  and  O  respectively. 

The  lines  of  action  of  these  reactions  are  Ox,  xO'y,  yO", 
drawn  parallel  to  DA,  DB,  and  DC  respectively.  These  lines 
should  intersect  on  ab  and  be  as  shown,  if  the  construction  is 
accurate. 

In  the  case  of  a  symmetrical  arch,  symmetrically  loaded,  the 
reactions  at  the  crown  will  evidently  be  horizontal.  The  re- 
actions of  the  end  hinges  can  then  for  this  case  be  found  directly; 
the  points  x  and  y  being  determined  by  the  intersections  of  a 
horizontal  line  through  O'  with  the  resultant  loads. 

Fig.  40  can  now  be  interpreted  as  follows:  The  point  D  can 
be  taken  to  be  the  pole,  and  the  lines  Ox,  xy,  yO"  the  strings  of  a 
funicular  polygon  for  the  given  loads.  The  reactions  of  ihe  hinges 
can,  then,  be  determined  by  constructing  a  funicular  polygon 
for  the  loads,  such  that  the  three  strings  will  pass  through  the 
hinges.  The  strings  will  be  the  lines  of  action  of  the  reactions, 
and  the  lengths  of  the  corresponding  rays  their  magnitudes. 

Instead  of  using  the  resultant  loads,  the  actual  loads  may  be 
employed  in  applying  this  method.  In  this  case  it  is  only  neces- 
sary to  construct  a  funicular  polygon  for  the  given  loads,  such 
that  the  three  limiting  strings  will  pass  through  the  three  hinges. 
(For  methods,  see  Arts.  23  and  24.) 


68 


GRAPHIC  STATICS. 


52.  Determination  of  Reactions.  ALGEBRAIC  SOLUTION.  In 
order  to  secure  greater  accuracy,  it  may  be  desired  to  determine 
by  calculation  the  position  of  the  pole  of  the  required  funicular 
polygon.  This  is  done  by  calculating  the  reaction  at  O1,  since 
the  ray  DB  (Fig.  40)  or  PD  (Fig.  41)  represents  this  reaction. 

Let  R}*R  be  the  equal  and  opposite  reactions  at  O'  (Fig.  41). 
These  are  resolved  into  horizontal  (H)  and  vertical  (V)  com- 
ponents. Taking  moments,  about  O,  of  the  forces  acting  on 
the  left  half,  we  have  H-b'-V -a' =  IW'xf,  in  which  W  repre- 
sents any  load,  and  yf  its  arm.  Similarly,  the  moments  about 
O"  of  the  forces  acting  on  the  right  half  give  the  equation 
H-bf+  V-a'  =  2Wx.  By  solving  these  two  equations,  we  deter- 
mine the  values  of  H  and  F,  and  consequently  R.  This  value 


w,' 


FIG.  41. 

of  R,  laid  off  in  the  proper  direction  from  the  point  D  of  the  force 
polygon,  locates  the  pole  P  of  the  required  funicular  polygon. 
PA  and  PG  are  then  the  reactions  at  O  and  O"  respectively. 


ROOF-TRUSSES. 


69 


It  is  to  be  noted  that  PAD  and  PDG  are  the  polygons  of  external 
forces  for  the  two  half- ribs. 

PROBLEM.     The  semicircular  arch  (Fig.  42)  hinged  as  shown, 
is  loaded  with  a  dead  load  of 
8000    pounds    uniformly    dis- 
tributed  over   the   roof,   and 
also    with    a    wind    load    of  I ,_,--' 
8000    pounds    on    the    right 
side.     Find   the   reactions   of 
the   hinges:     (i)    graphically; 
(2)     by     calculation    as    ex- 
plained.     Also  draw  a  funic-  FIG.  42. 

ular  polygon  for  these  loads,  to  pass  through  the  three  hinges. 

53.  Determination  of  Stresses  in  Braced  Arches.  The 
determination  of  the  stresses  in  three-hinged  braced  arches  will 
be  illustrated  by  the  following  example.  Fig.  3,  Plate  II,  rep- 
resents half  of  a  symmetrical  three-hinged  arch.  The  outer  in- 
clined chord  is  divided  by  the  joints  into  6  divisions  of  5  ft.  each 
and  one  of  8.1  ft.  The  member  7,  4  is  drawn  so  as  to  make  the 
adjacent  segments  of  the  inner  chord  equal,  the  other  correspond- 
ing members  being  perpendicular  to  the  outer  chord.  The  arch  is 
loaded  with  a  dead  load  uniformly  distributed  over  the  outer 
chord,  together  with  a  wind  load  on  the  right  side.  The  wind 
pressure  on  the  vertical  sides  of  the  building  is  not  supported  by 
the  arches.  It  is  required  to  determine  the  stresses  in  the  wind- 
ward rib.  The  diagonals  are  assumed  to  be  tension  members. 
The  joint  loads  (pounds)  are  as  follows: 


Joint. 

ab 

be    cd 

dt 

ef 

fg 

gk 

h 

Total. 

Dead  Load  
Wind  Load  

1620 
2430 

2620  2000 
3930  3000 

2OOO 
3000 

2OOO 
3OOO 

200O 
3000 

2000 
3000 

1000 

1500 

15240 
22860 

?o  GRAPHIC  STATICS. 

1.  Determination  oj  Reactions  of  Hinges.     The  resultant  dead 
and  wind  loads,  IK  and  KA   (Fig.  3  A),  are  plotted  and  their 
resultant  (I A]  is  subdivided  into  the  joint  loads.     The  H  and  V 
reactions  of  the  middle  hinge,  calculated  by  the  method  of  Art.  52, 
are:   H=  14,860  Ibs.;    V  =  ggoo  Ibs.     These  are  plotted  (see  Fig. 
3  A),  thus  locating  the  pole  P.     A  funicular  polygon  for  the  re- 
sultant dead  and  wind  loads  is  now  constructed  by  drawing  the 
string  (a)  through  the  end  hinge,  parallel  to  PA,  and  the  string 
(i)  through  the  middle  hinge,  parallel  to  PI;   when  it  is  found 
that  the  closing  string  (k)  is  parallel  to  PK,  thus  checking  the 
location  of  the  pole.     The  reactions  of  the  hinges  are  thus  PI 
and  AP,  and  the  polygon  of  external  forces  is  PIHGFEDCBAP. 

2.  Line  of  Pressure.     The  funicular  polygon  for  the  resultant 
joint  loads  is  now  drawn  (see  Fig.  3).     The  significance  of  this 
polygon  should  be  noted.     Beginning  at  the  left  end,  the  string 
i  is  the  reaction  on  the  hinge,  its  magnitude  being  PI;  the  string 
h  is  the  line  of  action  of  the  resultant  of  this  reaction  and  the  load 
at  the  apex,  its  magnitude  being  PH;  the  string  g  is  the  line  of 
action  of  the  resultant  of  the  reaction  and  the  loads  ih  and  hg, 
its  magnitude  being  PG,  etc.     Thus  it  is  seen  that  any  string  of 
the  funicular  polygon  for  the  joint  loads,  drawn  through  the  hinges, 
represents  the  line  of  action  of  the  resultant  external  force  acting 
on  the  portion  of  the  rib  to  either  side  of  the  panel  in  question;  the 
magnitude  of  this  force  being  represented  by  the  corresponding  ray. 

This  funicular  polygon  is  called  the  Line  of  Pressure,,  since  its 
strings  represent  the  lines  of  action  of  the  resultant  pressure  on 
the  corresponding  panels  of  the  arch. 

3.  Determination  as  to  -which  Diagonals  are  Stressed.     The 
methods  of  Art.  45  are  employed  to  determine  which  diagonals 
are  stressed,  the  application  of  these  methods  being  based  on  the 
preceding  paragraph. 

Take,  for  example,  panel  g,  where  the  chords  are  parallel. 


ROOF-TRUSSES.  7r 

The  resultant  external  force,  R,  to  the  left  of  the  section  mn  is 
PG,  acting  along  the  string  g  as  shown.  The  component  of  this 
force  at  right  angles  to  the  chords  evidently  acts  upwards;  the 
corresponding  component  of  the  diagonal  stress  must  then  act 
downwards  (method  of  shears) ;  hence  if  the  diagonals  are  tension, 
members,  the  one  drawn  full  (16-17)  is  the  one  stressed. 

Again,  take  the  panel  c,  where  the  chords  are  not  parallel* 
The  resultant  external  force,  R' ',  to  the  left  of  m'ri ',  is  PC  acting 
along  the  string  c  as  shown.  The  moment  of  this  force  about 
the  intersection  (not  shown)  of  the  chord  members  is  evidently 
right-handed;  hence  the  moment  of  the  diagonal  stress  must 
be  left-handed;  therefore,  the  diagonals  being  tension  members, 
the  one  drawn  full  (8-9)  is  stressed.  In  Fig.  3  the  diagonals 
stressed  are  drawn  in  full  lines. 

4.  Calculation  of  Stress  in  c-8  by  Moments.    Having  previously 
found  8-9  to  be  the  diagonal  stressed  in  this  panel,  we  take  the 
moment  axis  at  the  intersection  of  this  diagonal  and  the  inner 
chord  member,  i.e.,  at  the  joint  o.     In  calculating  the  resultant 
moment  of  the  external  forces  to  the  left  of  m'n',  about  o,  we  may 
use  the  H  and  V  component  reactions  at  the  middle  hinge  and 
the  stated  joint  loads  to  the  left  of  m'n' ,  or  we  may  take  the 
moment  of  their  resultant  as  represented  by  the  string  c  and  the 
ray  PC.     In  the  latter  case  we  would  find  the  magnitude  of  this 
resultant  by  scaling  off  PC  from  the  stress  diagram,  and  the 
moment   arm  by  scaling  off  the  perpendicular  distance  from  o 
to   the   string   c.     This   moment   divided   by   the   perpendicular 
distance  from  o  to  c-8  would  give  the  stress  in  c-8.     Its  value 
is  35250  Ibs.  compression. 

5.  Completion  of  Stress  Diagram.     We  plot   the   computed 
stress  in  c-8  and  complete  the  diagram  in  the  usual  way,  check- 
ing jt  on  the  members  4-5  and  12-14  (marked  ck  in  the  stress 
diagram).     In  an  arch  of  many  members  more  sections  than  one 


72  GRAPHIC  STATICS. 

should  be  taken  to  secure  sufficient  accuracy.  In  the  preceding 
example  counterbracing  was  introduced  in  order  to  indicate  the 
use  of  the  line  of  pressure  in  determining  which  diagonals  are 
stressed.  It  is  advisable  to  draw  the  line  of  pressure  in  any 
case. 

PROBLEM.  Determine  the  stresses  in  the  leeward  rib  of  the 
arch  of  the  preceding  example,  also  the  stresses  for  dead  load 
alone.  Find  the  maximum  and  minimum  stresses  in  all  the  mem- 
bers for  dead  and  wind  loads. 

Figure  43  represents  one  of  the  arches  supporting  the  train- 
shed  roof  of  the  Philadelphia  and  Reading  Terminal  Railway. 
The  diagonals  are  tension  members.1 


FIG.  43. 

54.  Three-hinged  Arch.  Solid  Ribs.  Determination  of 
Stresses.  Let  Fig.  44  represent  such  an  arch,  the  line  of  pressure 
and  the  force  polygon  for  the  given  system  of  loads  being  drawn. 
The  resultant  external  force  acting  to  the  right  of  any  cross- 
section  n  is  DP  =  F,  its  line  of  action  being  the  corresponding 
string.  Apply  at  n,  the  centre  of  gravity  of  the  cross- section, 
opposite  forces  equal  and  parallel  to  F,  and  resolve  one  of  these 
forces  into  F'  =  Dx  and  F"=xP,  respectively  perpendicular  and 
parallel  to  the  section.  The  original  force  and  —  F  form  a  couple 
which  causes  a  bending  stress  at  the  section,  the  bending  moment 
being  equal  to  the  moment  of  the  original  jorce  about  the  centre  oj 

1  See  Trans.  Am.  Soc.  C.  E.,  Aug.,  1895. 


ROOF-  TRUSSES. 


73 


gravity  of  the  section.     In  addition,  P  is  the  direct  compression 

stress  and  F"  the  shearing  stress  at  the  section. 

We  thus  see  that  the  stress  at  the  section  is  made  up  of: 

i.  A  bending  stress,  the  bending  moment   (Af)  being    the 

moment  of  the  resultant  external  force  on  either  side  of  the  section, 

taken  about  the  centre  of  gravity  of  the  section.     If  we  let  /= 


*> 


FIG.  44. 


moment  of  inertia  of  the  section  about  the  neutral  axis  through 
the  centre  of  gravity,  and  y  the  distance  from  this  neutral  axis  to 
the  most  compressed  fibre,  we  have  for  the  maximum  intensity 
of  the  bending  (compression)  stress 


M 


(i) 


2.  A    uniformly    distributed    compression    stress,    F'  =  Dx, 
whose  intensity  is 


F' 


(2) 


74 


GRAPHIC  STATICS. 


A  =  area   of  section.     Combining    (i)    and    (2),   the  maximum 
intensity  of  the  compression  stress  will  be 


_ 

A 


(3) 


3.  A  transverse   shearing   stress,  F"  =  xP.     This  last  stress 
will  usually  be  comparatively  small  in  the  case  of  arches. 

55.  Bending  Moments  Proportional  to  Vertical  Intercepts. 
In  the  case  of  vertical  loads,  the  bending  moments  are  propor- 
tional to  the  vertical  intercepts  between  the  line  of  pressure  and 
the  centre  line  of  the  arch  ring. 

Proof:  Let  RS  (Fig.  45)  be  a  portion  of  the  centre  line  of  an 

arch,  the  corresponding  por- 
tions of  the  force  diagram  and 
A  line  of  pressure  being  shown. 
Let  n  be  any  section  of  the 
arch,  nx  being  drawn  perpen- 
dicular to  the  string  b,  and  ny 

/£  P^ z   vertical.      The  triangles   nxy 

and  PzB  are   similar,  hence 


But  PB-nx  =  bending  moment  at  n  (see  Art.  54),  hence  Pz-ny 
also  equals  the  bending  moment  at  n.  As  the  pole  distance 
(Pz)  is  constant  for  vertical  loads,  the  moments  are  propor- 
tional to  ny,  the  vertical  intercepts.  Q.E.D. 

PROBLEM.  Given  a  symmetrical  three-hinged  semi-circular 
arch  of  20  ft.  span  between  end  hinges.  The  hinges  intersect 
the  centre  line  of  the  arch  ring.  The  arch  is  loaded  with  a  ver- 
tical load  of  10,000  Ibs.  5  ft.  to  the  left  of  the  crown  hinge,  and 
with  a  vertical  load  of  2000  Ibs.  6  ft.  to  the  right  of  the  crown 


ROOF-TRUSSES.  75 

hinge.  The  cross- section  is  an  I  section  of  equal  flanges,  10 
inches  deep,  5  sq.  inches  section,  moment  of  inertia  =150  (inches). 
Find  the  maximum  intensity  of  the  compression  stress. 


§  6.  Bending  Stresses.     Sway-bracing. 

56.  Conditions  under  which  Bending  Stresses  occur.  In 
order  that  no  bending  stresses  may  occur  in  the  members  of  a 
frame,  the  following  conditions  must,  in  general,  be  fulfilled: 
i.  The  centre  lines  of  the  members  must  be  straight  and  inter- 
sect at  a  common  point  at  the  various  joints.  2.  External  forces, 
must  not  act  at  any  other  points  than  the  joints.  3.  The  various, 
members  must  be  free  to  turn  at  the  joints.  If  these  conditions 
are  not  fulfilled,  bending  stresses  of  greater  or  less  magnitude 
and  definiteness  will  occur  and  should  be  taken  into  account. 

Some  of  the  more  prominent  cases  in  which  such  bending; 
stresses  occur  are: 

i.  Curved  members.  2.  External  forces  acting  at  other' 
points  than  the  joints.  E.g.,  when  the  purlins  are  supported  on 
the  upper  chord  of  a  roof-truss  at  other  points  than  the  joints. 
3.  Incomplete  or  defective  trusses,  i.e.,  trusses  in  which  the  mem- 
bers are  either  insufficient  in  number  or  improperly  placed,  SO' 
that  their  resistances  to  tension  and  compression  are  inadequate.- 
to  secure  stability.  4.  Columns  in  buildings.  5.  Centre  lines 
of  members  forming  a  joint,  not  intersecting  at  a  common  point, 
etc. 

Maximum  Intensity  of  Stress.  The  determination  of  the 
maximum  intensity  of  the  combined  bending  and  direct  (tension 
or  compression)  stress  is  similar  in  all  cases  to  that  given  in  Art.  55. 
If  we  let  P  =  total  direct  stress  at  any  section  of  a  member,  A  = 
area  of  section,  M  =  bending  moment  at  section,  7  =  moment  of 
inertia  of  section  about  neutral  axis  through  centre  of  gravity, 


76 


GRAPHIC  STATICS. 


and  y  =  distance  from  this  neutral  axis  to  the  most  stressed  fibre; 
we  have,  for  the  maximum  intensity  of  the  combined  stress, 

P      M 


The  following  Articles  give  solutions  of  some  special  cases 
in  which  bending  stresses  occur: 

57.   Purlins    supported    at    Other    Points    than    the  Joints. 

EXAMPLE.  The  truss,  Fig.  46  A,  is  loaded  with  a  uniformly 
distributed  vertical  load  of  9000  Ibs.  The  purlins  divide  the  upper 
chord  into  six  equal  parts;  thus  the  load  at  each  intermediate 


CD) 


FIG-  46. 

purlin  point  is  1500  Ibs.,  and  at  each  end  point  750  Ibs.  It  is 
required  to  determine  the  stresses  in  the  frame,  including  the 
maximum  intensity  of  compression  in  the  different  segments  of 
the  upper  chord,  its  cross-section  being  8"  wide  by  10"  deep. 

Consider  the  portion  AB  of  the  upper  chord  and  the  load  at 
n.  The  components  of  this  load  at  A  and  B  are  500  Ibs.  and 
1000  Ibs.  respectively.  Therefore  apply  at  these  points  equal 
and  opposite  forces  of  500  Ibs.  and  1000  Ibs.  as  shown  in  Fig.  46  C. 
These  added  forces  will  neutralize  each  other  and  hence  not 


ROOF-TRUSSES.  •  77 

affect  the  stresses,  so  that  the  system  of  forces  given  in  Fig.  46  C 
are  the  equivalent  of  the  1500  Ibs.  load  at  n.  This  system  of 
forces  may  be  separated  into  two  systems,  namely:  i.  The  down- 
ward forces  of  500  Ibs.  and  1000  Ibs.  at  A  and  B  respectively. 
These  cause  direct  stresses  in  the  members  of  the  truss.  2.  The 
load  at  n  and  the  upward  forces  at  A  and  B,  these  latter  cor- 
responding to  the  reactions  of  a  beam  AB,  loaded  at  n.  This, 
system  of  forces  (shown  by  full  lines  in  Fig.  46  C)  will  cause 
bending  stresses,  also  direct  compression  and  shearing  stresses 
in  AB. 

The  preceding  analysis  leads  to  the  following  solution:  (i; 
Resolve  each  load  into  components  at  the  adjacent  joints,  thus 
obtaining  the  joint  loads  of  Fig.  46  B.  Determine  the  direct 
stresses  in  the  members  of  the  truss,  due  to  these  joint  loads,  by 
constructing  a  stress  diagram  in  the  usual  way. 

(2)  Treat  each  segment  of  the  upper  chord  as  an  oblique  beam 
supported  at  the  ends  (joints)  and  loaded  at  the  purlin  points, 
calculating  the  beam  stresses  in  the  usual  way. 

(3)  Combine  the  stresses  found  in  (i)  and  (2). 

(The  following  conception  of  the  case  may  be  of  service  in 
making  this  method  of  solution  clearer.  If  the  loads  of  Fig.  46  A 
were  supported  as  shown  in  Fig.  46  D,  instead  of  being  applied 
directly  to  the  upper  chord,  there  would  be  bending  stresses  in  the 
jack-rafters,  and  direct  stresses  only,  in  the  upper  chord.  In 
this  case  the  forces  acting  at  the  joints  of  the  upper  chord  and 
those  acting  on  the  jack- rafters,  together  with  the  resulting  stresses 
in  these  members,  would  be  determined  as  indicated  in  (i)  and 
(2)  respectively. 

In  Fig.  46  A  the  upper  chord  alone  serves  the  same  purpose 
as  the  jack- rafters  and  upper  chord  together  in  Fig.  46  D,  so  that 
if  the  cross- sections  of  jack-rafters  and  upper  chord  of  Fig.  46  D 
were  each  equal  to  the  cross-section  of  the  upper  chord  of 


78  GRAPHIC  STATICS. 

Fig.  46  A,  the  stresses  in  the  latter  would  be  equal  to  those 
found  by  combining  the  stresses  existing  in  both  jack-rafters 
and  upper  chord  of  Fig.  46  D. 

The  case  may  be  stated  otherwise  as  follows  :  Combining 
the  external  forces  acting  on  the  jack- rafters  and  upper  chord  of 
Fig.  46  D,  we  obtain  those  given  in  Fig.  46  A.  Therefore,  if  we 
combine  the  stresses  in  the  jack-rafters  and  upper  chord  of  Fig. 
46  D,  we  will  obtain  those  existing  in  the  upper  chord  of  Fig. 
46  A,  provided,  of  course,  that  the  cross-sections  are  identical 
in  each  instance.) 

Solution,  i.  By  drawing  a  stress  diagram  (not  shown)  for 
the  joint  loads  of  Fig.  46  B,  we  find  the  stress  in  AB  to  be  5850 
Ibs.  compression,  and  in  BC  to  be  4050  Ibs.  compression. 

2.  Treating  AB  (see  Fig.  46  C)  as  a  beam  loaded  with  1500 
Ibs.  at  n,  we  find  that  the  maximum  bending  moment  (at  n)  = 
500X5X12  =  30,000  inch-lbs.     Also,  by  resolving  the  reaction 
at  A  into  components  along  and  at  right  angles  to  the  beam,  we 
find  an  additional  direct  compression  stress  of  278  Ibs.  in  the  part 
of  AB  below  n,  due  to  the  obliquity  of  the  loading. 

3.  We  thus  find  the  maximum  direct  stress  (compression)  in 
AB  to  be   5850+278  =  6128   Ibs.,   and   the  maximum  bending 
moment  to  be  30,000  inch-lbs.     Substituting  these  values  in  the 
formula  of  Art.  56,  we  have,  for  the  maximum  intensity  of  com- 

6128     30,000-5 
pression  in  AB,  /""gT"  +     66        =  302  Ibs.  per  square  inch. 

In  a  similar  manner  we  find  the  maximum  intensity  of  com- 
pression in  BC  to  be  283  Ibs.  per  square  inch.  , 

It  will  be  observed  that  the  additional  stress  due  to  the  ob- 
liquity of  the  loading  is  small. 

PROBLEM.  Given  a  triangular  truss  of  20  ft.  span  and  6  ft. 
rise,  loaded  with  a  uniformly  distributed  vertical  load  of  8000  Ibs. 
applied  directly  to  the  rafters.  The  rafters  are  wooden  beams 


ROOF-TRUSSES.  79 

4  in.  by  10  in.  section.     Find  the  maximum  compression  stress 
per  square  inch  in  these  members. 


58.  Trussed  Beam.  EXAMPLE.  The  trussed  beam  DE 
(Fig.  47  A)  is  loaded  uniformly.  It  is  required  to  find  the  stresses 
in  the  members. 

As  a  beam,  DE  is  supported  at  three  equidistant  points  and 
loaded  uniformly.  The  reaction  of  the  middle  support  in  such  a 
case  (see  Strength  of  Materials)  is  %W,  and  of  each  end  support 
TW-  Therefore  apply  ^  6/8w  ^ 

equal  and  opposite  forces  of  ''|{  ^1  i  i  i  i  1  ill  1 1 1  liiijj^ 
%W    at     the     middle     and   D[%  w"--~J~  jVtW 


at  each  end  as  shown     :3/i6w    (A)       54^  :3/">w 

by    the    dotted    lines.    The     ,  ,  t  ,  j  t  t  ,  t  n  1  t  t 

forces  of  Fig.  47  A  can  now     1  \ 

j  .  3/16  w    (B)       Vsw 

be  separated  into  two  bal- 
anced systems,  shown  in 
Figs.  47  B  and  47  C  re- 
spectively.  Fig.  47  B  rep- 
resents  the  forces  acting  on  FIG.  47. 

DE  as  a  beam,  the  maximum  fibre  stress  being  found  by  the 
methods  of  "  Strength  of  Materials."  Fig.  47  C  represents  the 
forces  acting  at  the  joints,  the  resulting  tension  and  compression 
stresses  being  found  by  constructing  a  stress  diagram.  Finally, 
the  maximum  intensity  of  the  stress  in  DE  is  found  by  combining 
the  stresses  obtained  from  Figs.  47  B  and  47  C  in  the  same  manner 
as  in  Art.  57. 

59.  Incomplete  Trusses.  Frames  of  this  character  are 
usually  subjected  to  bending  as  well  as  direct  stresses.  The 
solution  of  a  simple  case  follows. 


8o 


GRAPHIC  STATICS. 


If  the  frame  of  Fig.  48  were  loaded  symmetrically  at  the  joints, 
no  bending  stresses  would  occur,  and  the  direct  stresses  would 
be  found  from  a  stress  diagram  in  the  usual  way. 


(A) 


(0 


FIG  48. 


Under  an  unsymmetrical  load,  however,  such  as  given  in 
Fig.  48,  the  rafters  would  evidently  deflect  equally  in  the  manner 
indicated  by  the  dotted  lines. 

In  order  to  determine  the  stress  in  AB,  resolve  the  load  at  A 
into  a  horizontal  component  P,  and  one  acting  along  the  rafter. 
We  thus  obtain  P  =  2000  tan  60°  =  3464  Ibs. 

Let  x  =  compression  stress  in  AB.  Now,  since  the  deflec- 
tions of  the  two  rafters  are  equal,  the  horizontal  pressures  at  A 
and  B  must  be  equal. 


/.  P—  x=x; 


=%P  =  1732  Ibs. 


The  frame  can  now  be  treated  as  a  triangular  truss  loaded 
at  A  with  2000  Ibs.  (vertical)  and  1.732  Ibs.  (horizontal)  acting 


ROOF-TRUSSES.  8  1 

to  the  left;  and  loaded  at  B  with  1732  Ibs.  (horizontal)  acting 
to  the  right,  in  addition  to  the  joint  loads  of  1000  Ibs.  each  at 
apex  and  left  support.  Following  the  method  of  Art.  57,  these 
loads  can  now  be  separated  into  the  systems  of  loads  shown  in 
Figs.  48  A,  B,  and  C. 

The  stresses  due  to  the  joint  loads  (Fig.  48  B)  are  deter- 
mined from  a  stress  diagram.  They  are  given  in  the  figure. 

Solving  Fig.  48  A,  we  find  additional  direct  stresses,  due  to 
the  obliquity  of  the  loads,  of  1250  Ibs.  compression  in  the  lower 
half,  and  1250  Ibs.  tension  in  the  upper  half  of  the  rafter.  We 
have  also  a  maximum  bending  moment  in  the  rafter  of  30000 
inch  Ibs. 

Solving  Fig.  48  C,  we  find  additional  direct  stresses  of  750 
Ibs.  compression  in  the  lower  half,  and  750  Ibs.  tension  in  the 
upper  half  of  this  rafter;  also  a  maximum  bending  moment 
of  30000  inch-lbs. 

.  Combining  the  stresses  in  these  three  diagrams,  we  obtain 
the  following  results: 

Max.  direct  stress  in  left-hand  rafter  =  3250  Ibs.  C  .....  (i) 

"         "         "      "  right-hand  rafter  =2  750  Ibs.  C.     ...  (2) 

Stress  in  lower  chord  =  2600  Ibs.  T  ..........  (3) 

Max.  bending  moment  in  left-hand  rafter  =  30000  in.-lbs.    .  (4) 

"  "  "        "  right-hand  rafter  =  30000  in.-lbs.    .  (5) 

Finally,  (the  rafters  are  assumed  to  be  4"  wide  by  10"  deep) 
we  have,  combining  (i)  with  (4)  and  .(2)  with  (5): 

Max.  intensity  of  compression  in  left-hand  rafter 


30000.5  _  _ 

40  n       333         " 


82  GRAPHIC  STATICS. 

Max.  intensity  of  compression  in  right-hand  rafter 


30000  -t; 
333      =  5'9  lbs<  per  Sq>  mch' 


60.  Trusses    supported    by    Columns.     (For    examples    see 
Fig.  33  G,  H,  and  P,  also  Figs.  49,  52,  and  54.)      The  entire 
frame,  including  columns,  will  be  treated  together,  the  points 
of  support  being  at  the  bases  of  the  columns. 

There  are  two  general  cases,  namely:  i.  When  the  columns 
are  rigidly  attached  to  substantial  foundations  so  that  they  may 
be  considered  fixed  in  direction  at  the  base.  2.  When  fastenings 
and  foundations  are  insufficient  for  this,  in  which  case  the  columns 
are  assumed  to  be  hinged  at  the  base. 

If  the  sides  of  the  building  are  masonry  of  sufficient  strength, 
the  frame  will  be  required  to  resist  wind  pressure  on  the  roof 
surface  only.  If,  however,  the  sides  are  of  wood,  corrugated 
iron,  etc.,  supported  by  the  columns,  the  frame  must  resist  the 
wind  pressure  acting  on  the  sides  as  well  as  roof  of  the 
building. 

Under  the  action  of  the  wind  or  other  non-vertical  loads  the 
columns  will  be  subjected  to  bending  stresses.  Special  cases  are 
discussed  in  the  following  articles. 

61.  Case  I.     Columns  Hinged  at  Base.     The  Joints  oj  the 
Frame  may  be  subjected  to  Any  System  of  Loads;  the  Columns,  in 
addition,    being   loaded    at    Other   Points   than   the   Joints.      In 
order  to  avoid  unnecessary  complication,  the  additional  column 
loading  will  be  assumed  to  be  a  horizontal  wind  load,  W,  uni- 
formly distributed   over  the  left-hand  column   (Fig.   49).     The 
method  used  in  deriving  the  formulas  which  follow  is  applicable 
to  any  other  system  of  column  loads. 


ROOF-TRUSSES. 


Let  Fig.  49  represent  the  frame,  knee-braces,  d,  },  and  d',  f 
connecting  the  columns  and 
truss.     The  joint  loads  are 
not    shown,    the    additional 
column     loading    (W)    only 


being  represented.     The  no-        ^+%r 


FIG.  49. 


tation  of  Art.  28  will  be  used 
as  far  as  it  is  applicable,  IH 
and  2V  representing  the 
horizontal  and  vertical  com- 
ponent loads  (including 
column  loading). 

Assumptions.  As  in  case  of  trusses  supported  on  masonry 
walls  (see  Art.  28),  this  solution  must  be  based  on  assumptions; 
those  which  appear  to  be  as  reasonable  as  any,  and  which  will 
be  employed  here,  being  that  the  truss  is  comparatively  rigid, 
the  distances  e-ef,  d-d' ,  and  b-bf  remaining  unchanged,  while  the 
columns  bend  as  shown  by  the  dotted  lines  e,  d,  bv  and  e1 ' ,  d',  b{, 
being  assumed  jree  to  turn  at  the  joints  e,  d,  b,  and  ef,df,  bf.  It 
follows  from  these  assumptions  that  the  deflections  of  the  two 
columns  at  d  and  d',  relative  to  straight  lines  joining  their  ends, 
will  be  equal 

Separation  oj  the  Loads  into:  i.  Those  causing  beam  stresses 
in  the  columns.  2.  Those  causing  direct  stresses  in  the  frame. 

In  order  to  make  clear  the  manner  of  separating  the  loads, 
we  will  employ  a  conception  similar  to  that  stated  in  Art.  57. 
It  is  as  follows:  The  left-hand  column  (Fig.  49)  is  united  to  the 
frame  at  the  points  e  and  d.  This  column  is  subjected  to  both 
direct  and  bending  stresses.  Now  let  us  imagine  that  a  separate 
piece  (Fig.  50  B),  of  the  same  cross-section  as  the  column,  and 
connected  with  the  frame  by  rods  e,  e  and  d,  d  be  employed  to 
resist  the  bending  stresses  (the  right-hand  column  being  simi- 


84 


GRAPHIC  STATICS. 


larly  treated),  so  that  the  columns  of  Fig.  50  A  will  be  subjected 
to  direct  stresses  only. 

Under  these  circumstances  the  forces  acting  in  Fig.  50  B 
will  include:  i.  The  horizontal  reaction  H^  at  the  base;  2.  The 
wind  pressure  W\  3.  The  reactions  ^  and  Sl  of  the  connecting 
rods.  These  forces  evidently  cause  bending  stresses.  Similar 


Rl  R 


—  —  » 
—  > 

9 
Sl         Si 

XT          \ 

:      Si         8«     . 

d  ' 

I 

I!    ™. 

-±-*- 

H#- 

FIG.  50. 

conditions  exist  at  the  right-hand  end,  except  that  there  is  no 
wind  load. 

The  forces  acting  on  the  frame  (Fig.  50  A)  will  include: 
i.  The  vertical  reactions  Vt  and  F2;  2.  The  joint  loads  whose 
H  and  V  components  are  represented  by  IH—  W  arid  IV  re- 
spectively; 3.  The  reactions  Rl}  St  and  R2,  S2  of  the  rods.  This 
loading  evidently  causes  direct  stresses  only,  in  the  members  of 
the  frame. 

If  now  we  combine  the  pieces  of  Figs.  50  B  and  50  C  with  the 
left-  and  right-hand  columns,  respectively,  of  Fig.  50  A,  the  equal 
and  opposite  reactions,  RI,  S1}  R2,  and  S2  will  neutralize  each  other 
so  that  the  combined  external  forces  will  be  identical  with  those 
of  Fig.  49.  This  being  the  case,  we  can,  by  combining  in  the 
usual  manner  the  stresses  existing  in  the  pieces  of  Figs.  50  B  and 


ROOF-TRUSSES.  85 

50  C  with  those  of  Fig.  50  A,  obtain  the  actual  stresses  in  the    \ 
frame. 

i.  Determination  of  the  Values  0}  R1}  R2,  S1}  S2,  Hl}  H2)  Vl}  and 
V2.     Applying  the  general  conditions  of  .  equilibrium   (IH  =  o; 
=  o)  to  the  forces  of  Figs.  50  A,  B,  and  C  we  have: 


FromFig.  50  B.    H^S^W-R^  (IH  =  o)     .     .     (i) 

Hi-c^^+S^c-a)          (IM  =  o)     .    .     (2) 

FromFig.  50  C.   H2  =  S2-R2  (2H  =  o)     .    .     (3) 


o)     .    .     (4) 

FromFig.  50  A.   S^S^R^R^+IH-W  (2H  =  o)     .    .     (5) 
Also:  Deflection  at  d  (Fig.  50  B)  =  deflection  at  d'  (Fig.  50  C)  .  (6) 

To  find  the  deflection  at  d,  we  have  from  the  ordinary  formulas 
for  deflection  of  beams,  treating  Fig.  50  B  as  a  beam  supported  at 
the  ends  and  loaded  with  Si  and  W: 

Deflection  at  d  due  to  Sl  =  ~ 


Deflection  at  d  due  to  W=     *  ^  .   (ca+ca— 
Adding,  we  have: 


Similarly: 

Total  deflection  at  ^  =  ^~~  ..........    (8) 


86  GRAPHIC  STATICS. 

Equating  (7)  and  (8)  and  reducing,  we  have 


W(c2+ca-a2) 

(d\ 

"2    "1         &a(c-a) 

Adding  (2)  and  (4)  and  substituting  H^H^—I^ 
rwe  have 
(2H)-c        W-c 

H  (Art.  28) 
/Tn\ 

21       c—  a        2(0—  a)    *    * 

Solving  (9)  and  (10),  we  have 
(IH}-c     W(c2+$ca-a*\ 

/TT\ 

01      2(0—  a)           i6a(c—a)     ' 
(IH}-c     W(c2-ya-a2) 

•    •    I11; 
/T0\ 

2     2(c-a)   *"       i6a(c-a)      '       ' 
Substituting  (n)  and  (12)  in  (2)  and  (4),  we  have 

•   •    v12; 

{T?\ 

1       2                i6ca 
IH     W(c2-3ca-a2) 

(TA\ 

Substituting  (n)  and  (13)  in  (i),  and  (12)  and  (14)  in  (3), 
we  have 


_ 
1 


2 


2(c-a~) 


i6c(c-a) 


.  (15) 


2(c-a)  i6c(c-a) 


(16) 


The  values  of  V^  and  V2  can  be  found  as  explained  in  Art.  28, 
using  either  the  entire  system  of  loads  or  the  external  forces  of 
•Fig.  50  A. 


ROOF-TRUSSES.  87 

<r^ 

Having  determined  the  above  quantities,  the  beam  stresses 
in  the  columns  are  calculated  from  the  loads  of  Figs.  50  B  and  C, 
while  the  direct  stresses  in  the  frame  can  be  found  by  drawing  a 
stress  diagram  for  the  joint  loads  (Fig.  50  A). 

62.  Case  II.  Columns  Hinged  at  Base  and  Loaded  Only  at 
the  Joints.  This  is  a  special  case  under  Case  I,  the  value  of  the 
loading  on  the  columns,  other  than  at  the  joints  (W  in  this  work), 
being  zero.  Substituting  W  =  o  in  Eqs.  (u),  (12),  (13),  (14),  (15), 
and  (16)  we  have 


Approximate  Solution  of  Case  I.  An  approximate  solution 
of  Case  I  may  be  made  by  substituting  for  the  intermediate  column 
loading  (W  in  this  work)  its  components  at  the  adjacent  joints, 
e  and  d,  or  d  and  b  (Fig.  49),  and  using  the  formulas  of  Case  II. 
The  error  involved  will  evidently  vary  with  each  individual  case 
(see  Art.  66).  -**&• 

63.  Case  III.    Columns  Fixed   at   the  Base.     The  Joints  of 
the  Frame  may  be  subjected  to  any  System  of  Loads;  the  Columns, 
in  addition,  being  loaded  at  Other  Points  than  the  Joints.    As  in      **f 
Case  I  the  intermediate  column  loading  will  be  taken  to  be  a 
uniformly  distributed  wind  load  W. 

Assumptions.  We  assume  as  before  that  the  truss  is  com- 
paratively rigid,  the  points  e,  d,  ef,  and  d'  deflecting  equally,  while 
the  columns  bend  as  shown  by  the  dotted  lines  edb2  and  e'd'b2 
of  Fig.  49. 


GRAPHIC  STATICS. 


""*  Separation  of  Loads.  Treating  the  columns  as  beams  fixed 
in  direction  at  the  base,  the  loading  is  separated  (see  Art.  61)  as 
shown  in  Fig.  51. 

Determination  of  Values  of  R1}  Sl}  R2,  and  S2.     It  follows  from 
the  assumptions  just  stated  that  the  deflections  of  the  columns 


sv 


'SH-W. 


d         «I  — 

ii       ^ 

t  !     (A) 

«—  1«        d' 

6 

Li_i. 

&' 

B) 

Vi 

« 

FIG.  51. 

at  e,  d,  ef,  and  d'  are  equal.     From  the  ordinary  formulas  for  slope 
and  deflection  of  a  beam  fixed  at  one  end  we  have 


(due 


l 
Deflection  at  *- 


(due  to  5,) 


(due  to  W7)   (due 

T^c3      R.c3 


Deflection  at 


(due  to  W) 


(due  to  52)  (due  to  R2) 

(due  to  52)          (due  to  R2) 


(due 


(3) 


Deflection  at  d'  =  ^rT-  ^J T-  ). 

6 


ROOF-TRUSSES.  89 

Equating  (i)  and  (3)  and  reducing,  we  have 

CS,-«^'-(*,-Jw--.    •  •   (5) 


Equating  (i)  and  (2)  and  reducing,  we  have 
2  =  *-*-  ~ 


*).  (6) 


Equating  (3)  and  (4)  and  reducing,  we  have 

2),    .    .........    (7) 


0 

From  2H  =  o,  Fig.  51  A,  we  have 

(S1+S2)-(R1  +  R2)=IH-W  ......    (8) 

From  equations  (5),  (6),  (7),  and  (8)  we  have  finally 


-^- 


„  2c2+2ca—  a2 
S2  =  R2  -  —2  -  ;  ..........    (u) 

S1  =  IH-W+(Rl+R2)-S2  ........    (12) 

From  eqs.  (9),  (10),  (n),and  (12)  the  values  of  Ru  R2)  S1} 
and  S2  can  be  readily  determined. 

The  beam  stresses  in  the  columns  and  the  direct  stresses  in 
all  the  members  of  the  frame  can  now  be  found  as  described  for 
Case  I. 

64.  Case  IV.  Columns  Fixed  at  Base  and  Loaded  Only  at  the 
Joints.  This  is  a  special  case  under  Case  III,  the  value  of  the 


9o 


GRAPHIC  STATICS. 


loading  on  the  columns,  other  than  at  the  joints  (W  in  this  work) 
being  zero.  Substituting  W  =  o  in  (9),  (10),  (n),  and  (12),  and 
reducing,  we  have 


4(c-a)(c+2fl) 


(13) 


(14) 


Approximate  Solution.  An  approximate  solution  of  Case  III 
may  be  made  as  described  for  Case  I. 

65.  Example.  Fig.  52  represents  an  intermediate  bent  of 
a  steel  framed  building,  the  sides  of  which  are  corrugated  iron 
attached  to  the  columns.  It  is  required  to  determine  the  stresses 
due  to  wind  pressure  on  the  left-hand  side  (columns  fixed  at  base). 


(B) 


(O 


The  loads  are  separated  as  explained  in  Art.  63  into:  i.  Those 
causing  beam  stresses  in  the  columns  (Fig.  52  B  and  C);  2. 
Those  causing  direct  stresses  in  the  frame  (Fig.  52  A). 

Distance  between  bents  =  16  ft. 

Assumed  wind  load  on  a  vertical  surface          =  30  Ibs.  per.  sq.  ft. 
Total  wind  load  on  column  =  W  =  19200  Ibs. 

Normal  wind  load  on  roof  (Hutton's  formula)  =  19  Ibs.  per  sq.  ft. 


ROOF-TRUSSES.  91 

Total  wind  load  (normal)  on  roof =P=  10336  Ibs. 
Horizontal  component  of  P  =  4864  Ibs. 

Vertical  component  of  P  =5=  9120  Ibs. 

IH=  19,200+4864  =24064  Ibs. 

a  =30  ft. 

c  =40  ft. 

Substituting  these  values  in  (9),  (10),  (u),  and  (12),  Art.  63, 
we  obtain 

^=10795  Ibs.;   .£2=11071  Ibs.;  ^  =  7457  Ibs.;    ,$2=19273^8. 

Figs.  52  B  and  52  C  can  now  be  solved  as  beams  fixed  at 
one  end,  with  the  following  results: 

Windward  column  (Fig.  526): 

Maximum  shearing  force  (at  base)      =    15862  Ibs. 
Maximum  bending  moment  (at  base)  =  175910  ft.-lbs. 

The  bending-moment  diagram  (see  Chap.  Ill)  is  shown  by 
dotted  lines. 

The  point  of  inflexion  (bending  moment  =  o)  is  distant 
^=14.1  ft.  from  base. 

Leeward  column  (Fig.  52  C) : 

Maximum  shearing  force  (upper  segment)  =  11071  Ibs. 
Shearing  force  in  lower  segment,  including  base=  8202  Ibs. 
Maximum  bending  moment  (at  base)  =  135350  ft.-lbs. 

The  bending-moment  diagram  (see  Chap.  Ill)  is  shown  by 
dotted  lines. 

The  point  of  inflexion  (bending  moment  =  o)  is  distant 
#,=  i6.s  ft.  from  base. 


92  GRAPHIC  STATICS. 

" 
Direct  stresses  in  frame  (Fig.  52  A)  : 

Taking  moments  about  base  of  right-hand  column,  we  have 


•'•  Ft=  1736  Ibs. 
and          V2  =  9120—  1736  =  7384  Ibs. 

The  values  of  Vl  and  F2  might  also  be  found  by  using  the 
actual  system  of  loads  on  the  frame  (Art.  28). 

The  direct  stresses  in  the  frame  can  now  be  found  by  con- 
structing a  stress  diagram  for  the  loads  of  Fig.  52  A.  Fig.  52  A 
is  suitably  lettered  for  this  purpose  and  the  external  force  polygon 
ABCDEFGHIJA  is  drawn. 

The  cross-section  of  the  columns  being  known,  the  bending 
stresses  can  be  computed  and  combined  with  the  direct  com- 
pression stresses  given  in  the  stress  diagram  (see  Art.  57). 

66.  Approximate  Solution  of  the  Example,  Art.  65.  (See 
Art.  64.)  Resolving  the  wind  load  on  the  column  into  compo- 
nents at  the  adjacent  joints,  we  obtain  (see  Fig.  53)  2400  Ibs. 
at  ev  9600  Ibs.  at  dv  and  7200  Ibs.  at  &,;  the  last  being  shown  in 
Fig.  53  B,  since  its  effect  is  to  cause  a  shearing  stress  at  the  base 
of  the  column  rather  than  direct  stresses  in  the  frame.  In  this 
case  2H  =  4864  +2400  +9600=  16864  Ibs. 

Substituting  in  (13)  and  (14),  Art.  64  we  obtain 

7^  =  £2  =1-383  Ibs. 
Sl  =  S2  =  19815  Ibs. 

Separating  the  loads  as  before,  we  obtain  the  systems  of  forces 
shown  in  Fig.  53  A,  B,  and  C.  Solving,  we  obtain  the  following 

results: 


ROOF-TRUSSES. 


93- 


Windward  column  (Fig.  53  B) : 

Maximum  shearing  force  (at  base)      =   15632  Ibs. 
Maximum  bending  moment  (at  base)  =  139130  ft.-lbs. 

The  point  of  inflexion  (bending  moment  =  o)  is  16.5  ft.  above 
the  base. 

Leeward  column  (Fig.  53  C) : 

Maximum  shearing  force  (upper  segment)  =  11383  Ibs. 
Shearing  force  in  lower  segment,  including  base=  8432  Ibs. 
Maximum  bending  moment  (at  base)  =139130  ft.-lbs. 

The  point  of  inflexion  is  16.5  ft.  above  base. 


R,+2400\ 


c 

ei/                             \ 

. 

S),      8^-9600 

U,o        at 

+£>**, 

A 

\F 

b, 

\* 

j 

J           H               I 

B) 

V,                                                  V2 

(C 

FIG.  53- 

Direct  Stresses   in  Frame    (Fig.    53   A).     Taking   moments 
about  base  of  right-hand  column,  we  have 


z-  9600)30-  (Rt+R2+  2400)40+9120-45-4864-48; 

/.     ^=1186  Ibs. 
and     F2  =  9i2o  —  F,  =  7934  Ibs. 


60^= 


The  polygon  of  external  forces  is  given  in  Fig.  53  A. 
A  comparison  of  the  stress  diagrams  (not  drawn)  for  Figs.  52  A 
and  53  A,  together  with  the  relative  values  of  maximum  shearing 


94 


GRAPHIC  STATICS. 


force  and  bending  moment  in  Arts.  65  and  66,  will  serve  to  in- 
dicate the  difference  in  the  results  obtained,  in  this  particular 
example,  by  the  two  methods  of  solution. 

When  such  frames  are  hinged  at  the  bases  of  the  columns 
the  stresses  in  general  are  larger  than  when  fixed  at  the  base. 

67.  Problems,  i.  Solve  the  example  of  Art.  65,  assuming 
the  columns  hinged  at  the  base.  Compare  the  results  with  those 
of  Art.  65 

2.  Solve  Problem  i  by  the  approximate  method,  and  compare 
results  with  those  of  Art.  66.     Also  compare  the  results  with  those 
of  Problem  i.  ' 

3.  Determine    the    maximum    bending    moment,    maximum 
shearing  force,  and  points  of  inflexion  for  the  columns  of  Fig.  54, 


r 


FIG.  54. 

due  to  a  vertical  load  of  20000  Ibs.  uniformly  distributed  over 
the  roof  surface,  together  with  a  normal  wind  load  of  12500  Ibs. 
on  the  inclined  roof  surface.  Also  determine  the  direct  stresses 
in  all  the  members  of  the  frame.  Assume  the  columns  to  be 
hinged  at  the  base. 


ROOF-TRUSSES. 


95 


c*. 


4.  Solve  Problem  3,  assuming  the  columns  to  be  fixed  at  the 
base. 


68.  Sway-bracing.  Illustration.  The  frame  (Fig.  55)  is 
made  up  of  a  number  of  transverse  frames  or  bents,  braced  so  as 
to  resist  distortion  of  any  kind.  Each  bent  is  composed  of  a 


SIDE  ELEVATION 


I    1 

IENDJ 


ELEVAT.ION 


FIG.  55. 

truss  supported  on  columns,  the  truss  and  columns  being  braced 
together  as  shown  in  the  end  elevation.  In  the  intermediate 
bents  knee-braces,  shown  by  the  dotted  lines,  are  commonly 
used;  while  at  the  ends  two  intermediate  columns  are  shown, 
the  bracing  for  these  consisting  of  diagonal  tension-rods  and 
horizontal  struts.  The  bents  are  braced  together  by  (i)  diagonal 
tension-rods  in  the  plane  of  the  upper  chords,  the  purlins  serving 
as  struts;  2)  diagonal  tension-rods  in  the  plane  of  the  lower 


96 


GRAPHIC  STATICS. 


chords;    (3)  diagonal  tension-rods  in  the  sides  of  the  building 
between  the  columns. 

Such  a  system  of  bracing  as  described  aims  to  prevent  dis- 
tortion of  the  individual  bents,  and  to  keep  the  various  bents  ver- 
tical and  in  line,  under  the  action  of  any  horizontal  force  such 
as  wind  on  the  side  or  end  of  the  building,  the  pull  of  belts, 
thrust  of  jib-cranes,  etc.  The  particular  system  of  bracing  em- 


FIG.  56. 

ployed  will  vary  with  the  circumstances.1    The  stresses  in  such 
bracing  can  usually  be  best  determined  by  algebraic  methods. 

The  following  example  indicates  a  suitable  solution  for  a 
simple  case,  the  system  of  bracing  shown  being,  however,  gen- 
erally inconvenient  to  employ  for  such  structures. 

69.  Example.  The  vertical  lines  (Fig.  56)  represent  two 
adjacent  tiers  of  columns  in  a  building  of  skeleton  construction. 
The  horizontal  lines  are  floor  girders  or  special  members  located 

^ee  Johnson's  "Theory  and  Practice  of  Modern  Framed  Structures"; 
also  Freitag's  "Architectural  Engineering." 


ROOF-TRUSSES.  97 

in  the  floor  system ;  these,  with  the  diagonal  tension-rods,  forming 
a  system  of  wind  bracing  for  a  width  of  side  wall  of  20  ft.  It 
is  required  to  find  the  stresses  resulting  from  a  wind  pressure  of 
30  Ibs.  per  sq.  ft.  on  the  side  of  the  building. 

The  wind  pressure  on  each  panel  is  30X12X20=7200  Ibs. 
The  resulting  pressures  at  the  joints  are  given.  Since  the  diagonals 
are  tension  members,  those  drawn  in  full  lines  will  be  stressed  for 
wind  on  the  side  shown  (see  §  4).  The  stresses  in  the  members 
cut  by  any  section,  xy,  may  be  found  algebraically  as  follows : 

Dealing  with  the  forces  acting  on  the  portion  of  the  frame 
above  the  section,  we  have  (i)  ^H  =  o. 

.'.  3600+7200  +  7200=^'  cos  6  =  .8F'. 
.'.  Fr  =  22500  Ibs.  (tension). 

(2)  Taking  moments  about  O, 

3600-24+7200- 12  =  i6F.     .'.  F=  10800  Ibs.  (tension). 

(3)  From  2M  =  o  or  2V=o  we  find  the  compression  stress 
in  the  leeward  column  to  be  24300  Ibs. 

To  find  the  stress  in  jk  we  have,  dealing  with  the  joint  Oy 
IH  =  o. 

.*.  Compression  stress  in  jk  =  F'  cos  6=  18000  Ibs. 
The  stress  diagram  for  this  example  is  given  in  Fig.  56. * 

1  See  Freitag's  "  Architectural  Engineering  "  for  discussion  of  sway-bracing 
in  steel  framed  buildings  of  skeleton  construction. 


CHAPTER  III. 

BEAMS. 

§  i.  Shearing  Force  and  Bending  Moment. 

70.  Definitions.     The  shearing  force  (S)  at  any  cross-section 
of  a  beam  is  the  measure  of  the  tendency  of  the  external  forces  to 
cause  the  portion  of  the  beam  lying  on  one  side  of  the  section  to 
slide  over  the  remaining  portion,  this  tendency  being  opposed 
by  the  resistance  to  transverse  shearing  of  the  material.     The 
value  of  -5  is  found  by  taking  the  algebraic  sum  of  the  external 
forces  lying  on  either  side  of  the  section. 

The  bending  moment  (M)  at  any  cross-section  of  a  beam  is 
the  measure  of  the  tendency  of  the  external  forces  to  cause  the 
portion  of  the  beam  lying  on  one  side  of  the  section  to  rotate  about 
the  section,  this  tendency  being  opposed  by  the  resistance  to 
longitudinal  tension  and  compression  of  the  material.  The  value 
of  M  is  found  by  taking  the  algebraic  sum  of  the  moments  of  the 
external  forces  lying  on  either  side  of  the  section,  about  the  neutral 
axis  of  the  section. 

71.  Graphical  Representation   of    5"    and    M.       The   beam 
(Fig.  57)  is  supported  at  the  ends  and  loaded  with  three  con- 
centrated loads.     These  loads,  AB,  BC,  and  CD,  are  plotted  to 
scale,  and  the  supporting  forces,  DE  and  EA,  are  determined 
by  constructing  the  funicular  polygon  (Fig.  57  A). 

In  the  S  diagram  (Fig.  57  B)  any  ordinate  from  XX',  e.g.,  n'm', 
represents  the  value  of  S  at  the  corresponding  section  of  the  beam. 


BEAMS. 


99 


These  ordinates  are  taken  from  the  external  force  polygon  and 
plotted  as  indicated.  Thus  at  any  section  in  space  b,  S=EA  — 
AB  =  EB,  etc.  The  following  points  should  be  noted:  i.  At 


FIG.  57. 

any  concentrated  load,  e.g.,  ab,  there  are  two  ordinates,  n"m\' 
and  ri'm" .  These  represent  the  values  of  5*  at  sections  imme- 
diately to  the  left  and  right  of  the  load,  respectively,  and  differ  in 
value  by  the  amount  of  the  load.  2.  5"  has  opposite  signs  on  the 
two  sides  of  the  load  be,  hence  passes  through  the  value  zero  at  this 
section.  3.  S  is  constant  for  any  unloaded  portion  of  a  beam, 
and  varies  uniformly  in  case  of  a  uniformly  distributed  load;  the 
corresponding  portions  of  the  -5*  diagram  consisting  of  horizontal 
and  inclined  straight  lines  respectively  (see  Figs.  57  to  62). 

M  Diagram.  It  has  been  shown  (Art.  20)  that  in  case  of  a 
beam,  any  intercept,  e.g.,  mn,  of  a  funicular  polygon  multiplied 
by  the  pole  distance,  y  (proper  attention  being  given  to  the  scales 
of  the  force  and  funicular  polygons),  is  the  value  of  M  at  the 
section  in  question.  Since  the  pole  distance  is  constant,  the 
intercepts  are  proportional  to  the  values  of  M.  A  funicular 
polygon  for  the  external  forces  acting  on  the  beam  therefore  con- 


ioo  GRAPHIC  STATICS. 

stitutes  the  M  diagram.  In  constructing  this  diagram  the  pole 
distance  should  be  taken  to  represent  a  convenient  number  of 
units,  e.g.,  1000  Ibs.  The  method  for  determining  the  strings 
between  which  any  intercept  is  measured  is  as  follows:  The 
value  of  M  at  any  section,  e.g.,  mn,  is,  by  definition,  equal  to 
the  algebraic  sum  of  the  moments  of  the  forces  (EA  and  AB] 
lying  to  one  side  of  the  section.  The  strings  in  question  are  the 
ones  between  which  these  forces  lie,  i.e.,  e  and  b  in  this  instance. 
It  should  be  noted  that  in  case  of  any  system  of  concentrated 
loads  the  M  diagram  is  made  up  of  straight  lines,  while  the  por- 
tion of  the  diagram  corresponding  to  a  uniformly  distributed 
load  is  a  parabola  which  can  be  constructed  as  explained  in 
Art.  13. 

Algebraic  Signs  0}  S  and  M.  The  following  arbitrary  rules 
are  in  vogue:  5  is  positive  if  the  tendency  is  for  the  left-hand 
portion  of  the  beam  to  slide  upwards  relatively  to  the  right-hand 
portion  and  conversely.  M  is  positive  if  the  tendency  is  for  the 
beam  to  bend  convex  downward  and  conversely  (see  Fig.  57). 
In  the  S  diagrams  positive  ordinates  will  be  plotted  upwards. 
In  the  M  diagrams  positive  intercepts  will  lie  above  the  closing 
strings  of  the  funicular  polygon  provided  that  the  pole  be  taken 
on  the  left  side  of  the  force  polygon  (see  Figs.  57  to  62). 

72.  Relation  between  S  and  M  Diagrams.  The  algebraic 
condition  for  maxima  and  minima  applied  to  bending  moments 

is  — r— =o.     It  is  proved  in  " Strength  of  Materials"  that  — =S. 
dx  dx 

Hence,  in  general,  the  sections  of  a  beam  at  which  the  M  diagram 
intercepts  are  greatest  (or  least)  correspond  with  those  where 
the  S  diagram  cuts  the  X  axis  (see  Figs.  57,  59,  61,  62).  This 
relation  between  the  two  diagrams  does  not  in  certain  cases 
(see  Fig.  58)  hold  true,  the  reason  being  that  the  numerically 
greatest  (or  least)  value  of  an  ordinate,  occurring  within  a  limited 


BEAMS.  10 1 

space,  is  not  necessarily  a  maximum  or  minimum  value  in  the 
mathematical  sense. 

73.  Relation  between  Moment  Diagram  and  Elastic   Curve. 
M  is  assumed  to  be  positive  where    the  beam    bends    convex 
downward  and  conversely,  so  that  the  general  shape  of  the  elastic 
curve  may  be  determined  by  observing  the  algebraic  sign  of  M 
in  the  various  parts  of  the  beam  as  indicated  in  the  M  diagram. 
The  sections  at  which  M  changes  sign  and  hence  equals  zero 
evidently  correspond  to  points  of  inflexion  of  the  elastic  curve. 
The  general  form  of  these  curves,  determ'ned  in  this  manner, 
for  Examples  4  and  5  (see  Art.  74),  are  represented  in  Figs.  61 
and  62  respectively,  the  points  of  inflexion  being  marked  with 
circles. 

The  following  examples  will  serve  to  indicate  suitable  methods 
for  constructing  the  5  and  M  diagrams. 

74.  Examples,     i.  The    cantilever,  Fig.   58,  is    loaded  with 
a  uniformly  distributed   load  (resultant  AB)  and  a  concentrated 
load  BC  as  shown.     The  5  diagram  (Fig.  58  B)  needs  no  ex- 


FIG.  58. 

planation.  The  M  diagram  (Fig.  58  A)  is  constructed  as  follows: 
Substitute  for  the  distributed  load  its  resultant  (represented  by 
the  dotted  arrow) ,  and  construct  the  funicular  polygon,  beginning 
at  the  free  end.  By  taking  the  pole  on  a  horizontal  line  through 


102  GRAPHIC  STATICS. 

C,  the  string  c  may  be  taken  to  coincide  with  the  beam;  the 
strings  b  and  a  are  then  drawn.  The  true  diagram  for  the  dis- 
tributed load  is  a  parabola  tangent  to  the  strings  a  and  b  at  the 
points  n  and  n',  corresponding  to  the  ends  of  the  load.  The 
parabola  is  constructed  as  explained  in  Art.  13. 

EXAMPLE  2.  The  beam  (Fig.  59)  is  supported  at  the  ends 
and  loaded  with  the  uniformly  distributed  load  shown.  The 
supporting  forces,  BC  and  CA,  are  determined  by  constructing 


a  funicular  polygon  for  the  resultant  load  AB.  The  actual  dia- 
gram for  the  distributed  load  is  a  parabola  tangent  to  the  strings 
a  and  b  at  n  and  n'  respectively.  The  true  funicular  polygon 
thus  formed,  shown  by  heavy  lines  in  Fig,  59  A,  constitutes  the 
M  diagram.  The  5  diagram,  Fig.  59  B,  needs  no  explanation. 

From  the  S  diagram  it  is  seen  that  S  is  zero  at  the  point  O, 
the  maximum  intercept  ST,  and  hence  maximum  M  occurring 
at  this  section  of  the  beam. 

EXAMPLE  3.  If,  in  case  of  a  distributed  load,  the  value  of  M 
at  some  designated  section  of  the  beam  is  wanted,  it  can  be  found, 
without  constructing  the  funicular  curve,  by  making  such  section 


BEAMS.  103 

a  point  of  division  of  the  load  (Art.  12).  Fig.  60  represents  a 
beam  supported  at  the  ends  and  loaded  uniformly.  It  is  re- 
quired to  find  the  value  of  M  at  the  middle  of  the  beam.  Making 


BD                    J 

d    •      ' 

'          /               S 

c    \'                  T 

^S.         i 

0    n                                              n! 

FIG.  60. 

this  section  a  point  of  division  of  the  load,  we  substitute  for  each 
half-load  the  resultants  AB  and  BC  and  construct  the  funicular 
polygon  shown.  The  ordinate  ST  multiplied  by  the  pole  dis- 
tance gives  the  value  of  M  at  the  section  in  question.  In  the 
true  M  diagram  a  parabola  tangent  to  the  strings  a,  b,  and  c  at,  JP 
n,  S,  and  n'  respectively,  would  take  the  place  of  these  strings.  \  °  ^—* 

EXAMPLE  4.  Fig.  6 1  represents  a  beam  overhanging  its  Jju^ 
right-hand  support  and  loaded  with  a  uniformly  distributed 
load  over  its  entire  length  together  with  two  concentrated  loads 
BC  and  DE.  In  place  of  the  two  segments  of  the  distributed 
load  we  substitute  their  resultants  AB  and  CD  and  construct 
the  funicular  polygon,  Fig.  61  A.  The  closing  string  /  and 
corresponding  ray  determine  the  supporting  forces  EF  and  FA~ 
A  parabola  tangent  to  the  strings  a  and  b  at  n  and  S  respectively 
is  the  true  diagram  for  the  left-hand  segment  of  the  distributed 
load,  while  another  parabola  tangent  to  the  strings  c  and  d  at  S 
and  n'  respectively  is  the  true  diagram  for  the  right-hand  seg- 
ment. The  M  diagram  is  drawn  in  heavy  lines. 

The  5  diagram  is  constructed  as  follows:  The  value  of  S 
at  the  left-hand  support  is  FA  (reaction  of  support).  This 
distance  is  taken  from  the  external  force  polygon  and  plotted  in 


104 


GRAPHIC  STATICS. 


the  5  diagram  (lettered  FA').  On  the  left-  and  right-hand 
sides  of  the  load  BC  the  values  of  61  are  FA-AB  =  FB  and 
FA-AB-BC=FC.  These  values  are  plotted  and  lettered 
F'B'  and  F'C'  respectively.  On  the  left-  and  right-hand  sides 
of  the  supporting  force  EF,  the  values  of  S  are  FA  —  AB—  BC—  CX 
=  FX  and  FA-AB-BC-CX+EF=EX  (lettered  F'X"  and 
E'Xf  respectively).  The  load  CX  is  the  portion  of  CD  lying 


FIG.  61. 

to  the  left  of  the  right-hand  support.  Finally,  the  value  of  S 
at  the  right  end  of  the  beam  is  DE.  S  varies  uniformly  between 
these  sections  (see  Art.  71). 

Max.  51  (F'X")  occurs  immediately  to  the  left  of  the  right- 
hand  support. 

The  maximum  values  of  M  (S=o)  occur  at  the  load  BC  and 
at  the  right-hand  support,  the  numerically  greater  value  being 
at  the  load  BC  as  is  seen  by  inspection  of  the  M  diagram.  From 
the  M  diagram  it  is  seen  that  M  =  o  at  the  point  O,  being  positive 
to  the  left  and  negative  to  the  right  of  this  section. 

The  genera]  form  of  the  elastic  curve  is  indicated  in  the  upper- 


BEAMS. 


105 


most  diagram,  its  point  of  inflexion,  marked  by  a  circle,  being 
at  the  section  O. 

EXAMPLE  5.  Fig.  62  represents  a  beam  made  up  of  two 
segments  hinged  together  as  indicated.  It  is  supported  at  three 
equidistant  points  and  loaded  uniformly,  the  resultant  loads  on 


FIG.  62. 

the  two  segments  being  AB  and  BC.  A  funicular  polygon  for 
these  resultant  loads  is  constructed  as  follows:  Since  there  can 
be  no  bending  moment  at  the  hinge,  the  intercept  between  the 
strings  b  and  d  must  be  zero  at  that  point.  Hence  the  construc- 
tion is  begun  by  drawing  the  string  b  (  =  xy)  in  any  convenient 
direction  and  the  corresponding  ray;  the  pole  P  being  taken 
on  this  rayi  The  strings  a  and  c  are  next  drawn.  The  string 
d  must  now  pass  through  the  points  n'  and  O,  intersecting  the 
middle  support  at  n" '.  This  determines  the  closing  string  e 
(nn"}.  Rays  parallel  to  the  strings  d  and  e  determine  the  sup- 
porting forces  CD,  DE,  and  EA. 

The    true    diagram  for  the  distributed  load  is  a  parabola 


106  GRAPHIC  STATICS. 

tangent  to  the  strings  a,  b,  and  c  at  the  points  w,  O,  and  «'  re- 
spectively. The  M  diagram  is  drawn  in  heavy  lines. 

The  5  diagram,  Fig.  62  B,  is  constructed  as  described  for 
Example  4,  AX  =  XC  =  \AC  being  the  resultant  loads  on  the 
two  sides  of  the  middle  support.  Max.  5  occurs  on  both  sides 
of  the  middle  support. 

Max.  M  (5  =  o)  occurs  at  K,  K',  and  the  middle  of  the 
beam,  the  greatest  value  being  at  the  middle  of  the  beam  as  shown 
by  inspection  of  the  M  diagram.  M  is  zero  at  O  arid  Of  (Fig.  62  A), 
these  points  being  consequently  the  points  of  inflexion  of  the 
elastic  curve.  This  curve  is  drawn,  one  point  of  inflexion  being 
at  the  hinge. 

75.  Problems,  i.  Given  a  beam,  16  ft.  span,  supported  at 
both  ends  and  loaded  with  a  uniformly  distributed  load  of  1000 
Ibs.  per  foot,  extending  over  the  left  half-span.  Construct  the 
S  and  M  diagrams  and  determine  the  values  of  max.  S  and  max. 
M  graphically.  Check  the  graphical  work  by  determining  these 
values  algebraically. 

PROBLEM  2.  Given  a  beam  supported  at  two  points  16  ft. 
apart  and  overhanging  the  right-hand  support  4  feet.  It  is 
loaded  with  a  uniformly  distributed  load  of  1000  Ibs.  per  ft. 
covering  the  left  half-span  (8  ft.),  also  with  a  concentrated  load 
of  6000  Ibs.  at  the  extreme  right  end  of  the  beam.  Construct 
the  S  and  M  diagrams  and  determine  the  values  of  max.  S  and 
max.  M  both  graphically  and  algebraically.  Also  locate  the 
point  of  inflexion  of  the  elastic  curve  both  graphically  and  alge- 
braically. 

PROBLEM  3.  Construct  the  5  and  M  diagrams  for  the  col- 
umns of  Fig.  52  and  determine  the  values  of  max.  5  and  max.  M 
for  each  graphically. 


BEAMS. 


107 


2.     Deflection  of  Beams. 

*~-  76.  Graphical  Determination  of  Elastic  Curve.  In  simple 
cases  the  usual  formulas  furnish  the  best  method  -for  determining 
the  deflection  of  a  beam.  In  the  case  of  beams  of  non-uniform 
section,  or  loaded  in  a  complex  manner,  a  graphical  solution 
can  be  employed  advantageously  where  extreme  accuracy  is  not 
required. 

EXAMPLE.  The  beam  (Fig.  63),  of  uniform  section,  is  sup- 
ported at  the  ends  and  loaded  with  two  concentrated  loads  as 
shown.  It  is  required  to  construct  the  elastic  curve. 

Solution.  Construct  the  funicular  polygon  (Fig.  63  A)  for  the 
given  loads.  The  force  diagram  is  Fig.  63  A'.  Treat  the  surface 


FIG.  63. 

of  this  polygon  as  if  it  represented  a  distributed  load  for  the  same 
beam,  and  construct  a  second  funicular  polygon.  For  this  pur- 
pose the  surface  (Fig.  63  A)  is  subdivided  by  ordinates,  and  each 
division  is  concentrated  at  its  centre  of  gravity,  marked  by  a 
circle.  The  areas  of  these  surfaces  are  plotted  to  scale  in  the 
force  diagram  (Fig.  63  B'),  and  the  corresponding  funicular  poly- 


io8  GRAPHIC  STATICS. 

*•—+. 
gon  (Fig.  63  B)  is  drawn.     The  exact  diagram  will  be  a  curve 

inscribed  in  this  polygon,  its  tangent  points  corresponding  to  the 
points  of  division  of  Fig.  63  A.  This  curve  is  the  elastic  curve 
of  the  beam,  the  deflection  at  any  point  being  represented  to  scale 
by  the  length  of  the  intercept  between  the  curve  and  the  line  SSf, 
this  line  being  drawn  so  as  to  satisfy  the  condition  that  the  deflec- 
tion at  each  support  is  zero. 
Proof: 

Let   O  =  any  ordinate  of  Fig.  63  A  (inches,  full  size). 
Z)  =  pole  distance  in  Fig.  63  A'  (Ibs.). 
Z>'  =  pole  distance  .in  Fig.  63  B'  (sq.  inches,  full  size). 
M  =  bending  moment  at  any  section  of  the  beam  (inch-lbs.). 
E  =  modulus  of  elasticity  of  material  (Ibs.  per  sq.  inch). 
/  =  moment  of  inertia  of  beam  section  (inches). 
Take  the  origin  of  coordinates  at  S,  X  axis,  horizontal,  and  Y 
axis,  vertical.     (For  convenience,  the  pole  P'  is  here  taken  so 
that  SS'  is  horizontal.)     Let  mn  be  tangent  to  the  elastic  curve 
at  any  point  C"  whose  coordinates  are  x,  y.    The  rays  P'S  and 
P'C'  are  parallel  respectively  to  SS1  and  mn.     Hence 

dy    SC     SA'-A'C 


In  Eq.  i,  SA'  is  constant  and  A'C'  is  equal  to  the  area  in  Fig. 
63  A,  lying  to  the  left  of  the  point  being  considered;  i.e., 

A'C'=   f*Odx. 

Jo 

Substituting  in  Eq.  i, 


dy_ 
dx 


SA'—  f*0d* 
Jo 


BEAMS.  109 

\, 

Differentiating  and  dividing  by  dx,  we  have,  neglecting  signs, 

£U£  (2) 

*~' (2> 


The  general  (approximate)  equation  of  the  elastic  curve  is 
d2y     M 


dx-~EI 


(3) 


Hence  the  curve  of  Fig.  63  B  will  be  the  true  elastic  curve 
provided  that  the  second  members  of  Eqs.  2  and  3  are  equal,  i.e., 


If  the  pole  distances  are  taken  at  random,  the  relative  lengths 
of  the  ordinates,  Fig.  63  B,  will  represent  the  relative  deflections 
at  the  various  points  of  the  beam.  If  D  and  Dr  are  taken  so  as 
to  satisfy  Eq.  4,  the  scales  of  abscissas  and  ordinates  will  be  the 
sarhe.  It  is  desirable  to  magnify  the  ordinates  so  as  to  determine 
the  deflections  more  accurately  than  otherwise.  Thus  if  the 
scale  of  abscissas  is  i  :  a  and  we  wish  to  make  the  ordinates  in  the 
drawing  n  times  as  great  as  the  actual  deflections,  we  must  make 
the  scale  of  ordinates  nY,a  as  great  as  the  scale  of  abscissas  in 
Fig.  63  B.  For  this  purpose  DD'  must  be  reduced  in  the  same 
ratio  (see  Art.  20)  and  Eq.  4  then  becomes 


By  means  of  Eq.  5  suitable  values  of  D  and  Df  can  be  selected 
such  that  the  ordinates  of  Fig.  63  B  will  represent  the  deflections 
of  the  beam  magnified  n  times. 

It  is  not  necessary,  in  determining  deflections,  for  SS'  (Fig. 
63  B)  to  be  horizontal,  since  the  ordinates  remain  the  same  so 
long  as  the  pole  distance  Dr  is  not  altered. 


no  GRAPHIC  STATICS. 

MM* 

If  the  lending  moment  changes  sign,  the  areas  (Fig.  63  A) 
corresponding  to  negative  moments  must  be  plotted  in  the  force 
polygon  (Fig.  63  B')  in  the  opposite  direction  from  those  corre- 
sponding to  positive  moments  (see  Example  2,  Art.  77). 

If  the  beam  is  of  non-unijorm  section,  Eq.  5  shows  that  DDr 
must  vary  in  the  same  ratio  as  /.  This  is  accomplished  by  varying 
the  pole  distance  D'  (see  Example  i,  Art.  77). 

To  determine  the  deflection  at  any  given  section  of  the  beam 
make  one  division  line  of  the  moment  diagram  (Fig.  63  A)  cor- 
respond to  such  section  (compare  with  Example  3,  Art.  74). 
Otherwise  the  moment  diagram  may  be  subdivided  in  any  man- 
ner, except  in  case  of  beams  of  non-uniform  section,  for  which 
see  Example  i,  Art.  77. 

77.  Examples,  i.  Fig.  i,  Plate  II,  is  a  cantilever,  10  ft. 
span,  loaded  at  two  points.  /  has  three  different  values,  as 
given  in  the  diagram.  £  =  30000000  Ibs.  per.  sq.  in.  It  is 
required  to  construct  the  elastic  curve  and  determine  the  deflection 
at  the  free  end  of  the  beam. 

Scale1  of  space  diagram,  1:20.  .'.a  =20.  Scale  of  force 
diagram  (Fig.  i  A'),  4000  lbs.  =  i".  The  pole  distance  PA  is 
taken  to  be  4".  .'.  .0  =  4X4000  =16000.  The  construction  of 
the  moment  diagram  (Fig.  i  A)  needs  no  explanation.  Its  surface 
is  divided  as  shown  and  the  centres  of  gravity  of  the  divisions 
are  indicated  by  circles.  The  areas  of  these  divisions  are  plotted  in 
Fig.  i  B'  to  the  scale,  1000  sq.  in.  (full  size)  =  i".  The  full-size 
area  is  obtained  by  multiplying  the  diagram  area  by  a2=(2o)2  = 
400.  The  deflections  are  to  be  magnified  five  times,  i.e.,  n=$. 

Substituting  the  preceding  values  in  the  formula  DD'  =  — , 

1  The  scales  referred  to  here  are  those  of  the  original  drawings,  which  have 
ibeen  reduced  about  one-half  in  the  plate. 


BEAMS.  : 

^OOOOOOO • 2OO 

we  have    16000  D' =  - — -,     .-.  £'  =  3750  =  3.75".    The 

pole  P'  is  taken  with  a  pole  distance  of  3-75",  and  the  elastic  curve 
(Fig.  i  B)  is  then  constructed  as  follows :  The  strings  correspond- 
ing to  P'Gf,  P'F',  P'E'  are  drawn  in  order.  At  the  section  e' 
the  value  of  I  changes  to  150,  and  the  pole  distance  must  be 
reduced  in  the  same  ratio,  the  new  pole  P"  lying  on  P'FJ .  The 
construction  is  continued  in  a  similar  manner,  the  pole  for  the 
portion  of  the  beam  where  7=ioo  being  P'".  A  curve  tangent 
to  these  strings  at  the  points  corresponding  to  the  points  of  division 
of  Fig.  i  A  is  the  elastic  curve.  The  ordinate  mn,  divided  by  5, 
determines  the  deflection  of  the  beam  at  the  free  end  to  be  .496". 
The  computed  deflection  is  .493" . 

The  following  points  concerning  Fig.  i  B  should  be  noted: 
i.  The  vertical  scale  is  5-20=  100  times  as  great  as  the  horizontal 
scale.  2.  In  order  to  determine  the  true  deflection  at  the  free 
encl,  it  would  have  been  sufficient  to  have  divided  Fig.  i  A  only 
at  the  two  sections  where  /  changes  in  value. 

If  the  value  of  /  varied  continuously,  the  divisions  of  the  beam 
and  moment  diagram  (corresponding  to  cf  and  e*,  where  /  changes) 
should  be  made  sufficiently  numerous  to  insure  a  close  approxi- 
mation to  the  true  result,  the  mean  value  of  /  for  each  portion 
of  the  beam  being  employed  in  the  construction  of  the  elastic 
curve.  Otherwise  the  solution  for  such  a  case  is  similar  to  that 
just  explained. 

EXAMPLE  2.  The  beam  (Fig.  2,  Plate  II)  is  supported  at 
two  points  a  and  h,  14  ft.  apart,  and  overhangs  6  ft.  at  the  right 
end.  It  is  loaded  with  a  distributed  load  of  300  Ibs.  per  foot. 
Cross-section  =  6" X 1 2".  7  =  864.  £=1200000.  Scale  of  space 
diagram,  i  :  40. 

The  load  is  divided  into  ten  equal  lengths,  each  division  being 
concentrated  at  its  middle  point.  Scale  of  force  diagram  (Fig.  2  A'), 


H2  GRAPHIC  STATICS. 

* 

2000  Ibs.  =  i".  Pole  distance  =  i  J";  i.e.,  D  =  ^ooo.  The  sur- 
face of  the  moment  diagram  1  (Fig.  2  A)  is  divided  as  shown  and 
the  areas  plotted  in  Fig.  2  B'  to  the  scale,  500  sq.  in.  =  i",  the 
positive  and  negative  areas  being  laid  off  in  opposite  directions. 
»=5- 

777" 

Substituting  in  the  formula  DD'  =  —  ,  we  have 


1200000-864 


'  =  1728. 


Hence  the  pole  distance  (Fig.  2  B')  is  -^W  =  3-456".  Fig.  2  B 
is  the  elastic  curve.  The  measured  ordinates  at  7  ft.  and  20  ft. 
from  the  left  end,  divided  by  5,  determine  the  deflections  at  these 
points  to  be  —.144"  and  +.012"  respectively. 

The  true  shape  of  the  elastic  curve  here  would  be  obtained 
by  plotting  its  ordinates  from  a  horizontal  line,  reducing  their 
scale  40  •  5  =  200  times. 

78.  Problem.     Given  a  beam  16  ft.  span,  supported  at  both 
ends  and  loaded    in    the    middle  with  4000  Ibs.,  £  =  30000000. 
In  the  middle  half  of  the  span  the  cross-section  is  uniform,  the 
value  of  /  being  300.     Outside  the  middle  half  the  value  of  / 
diminishes  at  a  uniform  rate  to  each  support  where  7=ioo.     It 
is  required  to  determine  the  deflection  at  the  middle  of  the  span. 

(Note.  In  constructing  the  elastic  curve,  divide  the  middle 
portion  of  the  span  into  two  4-  ft.  segments  and  each  end  portion 
into  four  i-ft.  segments,  using  the  average  value  of  /  in  each.) 

% 

79.  Centre  of  Gravity.     The  following  constructions  for  centre 
of  gravity  will  be  found  useful  in  connection  with  the  work  of  this 
and  the  following  chapter. 

i.  Centre  o]  Gravity  oj  Any  Quadrilateral  Area.  Let  A  BCD 
(Fig.  64)  be  the  quadrilateral  in  question.  Draw  the  two  diago- 

1This  diagram  might  better  have  been  constructed  by  the  method  of  Art.  74. 


BEAMS, 


nals.  Bisect  one  of  these,  as  AC,  at  E.  Also  lay  off  DK'  =  BK 
and  draw  EK' .  Trisect  EK'  at  G.  G  is  the  centre  of  gravity 
of  the  area. 


FIG.  64. 


FIG.  65. 


2.  Centre  of  Gravity  of  a  Trapezoid.  The  following  special 
construction  is  useful:  Let  ABCD  (Fig.  65)  be  the  trapezoid. 
Bisect  each  of  the  parallel  sides  and  draw  the  medial  line  EEr. 
Extend  the  two  parallel  sides  in  opposite  directions,  laying  off 
AF=BC  and  CF  =  AD.  Draw  FF',  intersecting  EEf  at  G. 
G  is  the  centre  of  gravity  desired. 


CHAPTER  IV. 

MASONRY  ARCHES,  ABUTMENTS,  ETC. 

§  i.     General  Conditions  oj  Stability. 

80.  Nature  of  the  Forces  involved.  Let  PRMN  (Fig.  66) 
be  a  block  of  masonry  acted  upon  by  a  force  AB.  In  addition 
to  this  force,  the  weight  of  the  block  must  be  taken  into  account. 
This  weight  is  represented  by  BC,  its  line  of  action  be  being 
drawn  through  the  centre  of  gravity  of  the  block.  The  resultant 
of  these  two  forces  is  AC,  its  line  of  action  passing  through  the 


M  N 

FIG.  66. 

intersection  of  ab  and  be.  AC  is,  therefore,  the  resultant  pressure 
exerted  by  the  block  upon  the  plane  MN.  This  plane  may  be 
taken  to  be  a  joint  of  the  masonry  or  its  base.  Also,  the  forces 
which  hold  the  block  in  equilibrium  are  AB,  BC,  and  the  reaction 
of  the  plane  MN,  this  last  being  a  force  equal  and  opposite  to 
AC.  Moreover,  AC  represents  the  resultant  stress  on  the  plane 
MN. 

Again,  let  PRMN  (Fig.  67}  be  an  arch-stone,  BC  being  the 

114 


M/tSONRY  ARCHES,  ABUTMENTS,  ETC.  "5 

load  supported  by  this  stone  including  its  own  weight.  The  line 
of  action  be  of  this  load  passes  through  its  centre  of  gravity.  BC 
is  balanced  by  the  forces  AB  and  CA  exerted  upon  PRMN  by 
the  adjacent  arch-stones.  These  three  forces  must,  therefore, 
form  a  triangle  as  shown,  and  their  lines  of  action  must  intersect 
at  the  same  point. 

81.  Resistance    of    a    Masonry    Joint.      The    conditions    of 
stability  for  the  block  of  Fig.  66  as  far  as  the  joint  MN  is  con- 
cerned are  evidently  the  following:    (i)  the  block  must  not  over- 
turn about  an  edge,  as  N;    (2)  it  must  not  slide  over  the  joint; 
(3)  the  material  of  the  stone  and  mortar  must  not  crush.    These 
three  conditions  will  be  discussed  in  turn. 

82.  Resistance    to    Overturning.     In    this    connection,    the 
tensile  strength  of  the  mortar  joint  is  commonly  neglected.     Then 
the  block  (Fig.  66)  would  evidently  overturn,  if  the  line  of  action 
of  the  resultant  force  AC  pierced  the  plane  MN  outside  of  the 
surface  of  the  joint.     The  moment  of  AC  about  N  as  moment  axis 
is  the  measure  of  the  resistance  to  overturning  about  this  edge, 
i.e.,  in  order  to  overturn  the  block  it  would  be  necessary  to  apply 
a  force  whose  moment  about  N  was  equal  to  that  of  AC,  but 
having  the  opposite  sign. 

83.  Resistance  to  Sliding.     Let  the  resultant  force  AC  (Fig.  66) 
be  resolved  into  components  parallel  and  perpendicular  to  the 
joint,  as  indicated.     The  normal  component  represents  the  direct 
pressure  on  the  joint,  while  the  parallel  component  tends  to  slide 
the  block  over  the  joint,  and  must  be  resisted  by  the  sliding 
friction  at  the  joint,  the  adhesion  between  the  stone  and  mor- 
tar being  neglected. 

Coefficient  of  Friction.    Let   P   (Fig.   68)   be   the   resultant 


n6  GRAPHIC  STATICS. 

pressure  of  the  block  on  the  plane  AB,  and  (f>  the  minimum 
angle  of  inclination  with  the  normal  at  which 
sliding  will  occur.  This  angle  </>  is  called  the 
angle  of  repose,  and  tan  <j>,  or  ratio  of  tangential 
to  normal  component  of  the  force,  is  called  the 
coefficient  of  friction.  In  the  case  of  masonry 
joints,  the  value  of  the  coefficient  of  friction  is  taken  to  be  from 
.4  to  .5.  In  order,  then,  for  sliding  not  to  occur,  the  resultant 
pressure  at  any  joint  must  make  with  the  normal  an  angle  less 
than  tan"1 .4. 

84.  Resistance    to    Crushing.      The    normal    component    of 
AC  (Fig.  66)  represents  the  resultant  compression  stress  at  the 
joint  MN.     This   stress  is  assumed  to  be  uniformly  varying. 
The  three  cases  whch  may  occur  are  repre- 
sented in  Fig.  69.     In  Fig.  69  A  the  stress  is 
distributed  over  the  whole  surface  of  the  joint, 
the  limiting  case  being  Fig.  69  B,  where  the 
intensity  of   stress    is    zero    at    one    edge    M. 
In  Fig.  69  C  the  pressure  is  distributed  over    M 
the  portion  XN  of  the  joint.     If  the  joint  were  ihTll 

capable  of  resisting  tension,  this  last  would  be  ^ilUlljlUj- 

the  case  where  the  stress  is  partly  tension  and 
partly    compression,    but,    assuming    the   joint 
incapable  of  resisting  tension,  the  portion  MX  is  without  stress 
and  tends  to  open. 

Case  oj  Rectangular  Joint.      In  Fig.  69  A  the  maximum  in- 
tensity of  compression  (at  N)  is  given  by  the  formula  /=~r+  ~jy 

A.      1 

(Art.  54),  in  which  R  =  resultant  normal  pressure  on  joint;  A  = 
area  of  surface  of  joint;  M  =  moment  of  R  about  centre  of  gravity 
of  this  surface  =  R  •  X^ ;  /  =  moment  of  inertia  of  surface  of  joint 


MASONRY  ARCHES,  ABUTMENTS,  ETC.          II? 

about    its    centre    of    gravity,  and   y=\d    (d=MN=  depth   of 
joint). 

In  Fig.  69  B  the  intensity  of  compression  at  M  is  zero,  hence 

T>  yl    f  XT     TT>   XT 

—  =  -j-y  =  -^-°.    .'.  ^T0=  $d,  or  7?  acts  at  \MN  from  2V.    Hence  : 

ln  order  for  the  pressure  on  a  rectangular  joint  to  be  distributed 
over  its  entire  surjace,  the  resultant  must  act  within  the  middle  third  < 
oLthe  depth  of  the  joint.  __  .  ___  __  _         j  ^ 

The  maximum  intensity  of  stress  (at  N}  in  Fig.  69  B  is  evi- 

dently double  the  average  stress,  i.e.,  —  :-.     Similarly,  in  Fig.  69  C, 

A. 

the  maximum  stress  is  double  the  average  stress  on  the  surface 


Case  0}  Circular  Joint.  To  determine  the  limits  within  which 
R  must  act  in  order  for  the  pressure  to  be  distributed  over  the 

whole  surface  of  the  joint,  we  have  as  before  ~i  —  ^fy—  —  ~A  —  -• 

A.      1  A.r 

.-.  X0=}r  (r.  =  radius  of  joint  surface).  Hence  in  this  case  the 
resultant  must  act  within  one-fourth  0}  the  radius  from  the  centre 
of  the  joint. 

For  other  forms  of  joint,  the  limits  within  which  R  must  act 
in  order  for  the  pressure  to  be  distributed  over  the  entire  area,  and 
the  value  of  the  maximum  intensity  of  stress  when  the  pressure  is 
so  distributed,  are  determined  in  the  same  manner  as  indicated 
above  for  rectangular  and  circular  surfaces. 

85.  Conditions  to  be  Satisfied  by  Masonry  Arches,  Abutments, 
etc.  The  following  conditions  respecting  the  joints  should,  in 
general,  be  fulfilled: 

i.  The  limits  within  which  the  resultant  pressure  acts  should 
be  such  that  the  pressure  will  be  distributed  over  the  entire  surface 
of  each  joint. 


n8 


GRAPHIC  STATICS. 


2.  The  direction  of  the  resultant  pressure  should  be  nearly 
at  right  angles  to  the  surface  of  the  joint. 

3.  The  maximum  intensity  of  compression  on  each  joint,  in- 
cluding the  base,  must  not  exceed  the  safe  compression  strength 
of  the  material. 

§  2.  Masonry  Arch.    Line  of  Pressure. 

86.  Definitions.     In  Fig.  70,  a  and  b  are  respectively  the  span 
and  rise  of  the  arch,  h  is  the  thickness  of  the  arch-ring,  also  the 
of  the  joints.     The  inner  and  outer  surfaces  of  the  arch- 


FIG.  70. 

ring  are  the  intrados  and  extrados  respectively,  these  terms  being 
also  applied  to  the  lines  cmc'  and  dm'd'.  The  highest  part  of  the 
arch  is  the  crown.  The  surfaces  cd  and  dd'  are  the  skew-backs. 
The  portions  of  the  arch-ring  between  the  crown  and  skew- 
backs  are  the  haunches. 

The  space  B,  outside  the  extrados  and  within  the  dotted  lines, 
is  the  spandrel.  The  masonry,  usually  with  horizontal  joints, 
lying  in  space  B  is  the  backing.  The  timber  frame  which  sup- 
ports the  arch  during  erection  is  the  centring. 

Arches  are  designated  according  to  the  form  of  the  intrados 
as  semicircular,  segmental,  elliptical,  pointed,  etc. 

87.  Line  of  Pressure  a  Funicular  Polygon.  Let  AB,  BC, 
CD,  etc.  (Fig.  71),  be  the  loads  supported  by  the  arch-stones.  If 


MASONRY  ARCHES,  ABUTMENTS,  ETC. 


119 


FIG.  71. 


the  pressure  at  any  joint,  as  a,  is  given  completely,  the  pressures 
at  the  other  joints  can  be  found  by 
the  triangle  of  forces  (Art.  80). 
Thus,  representing  the  pressure  on 
the  joint  a  by  PA,  the  resultant  of. 
PA  and  AB,  i.e.,  PB,  will  be  the 
pressure  on  the  joint  b,  its  line  of 
action  passing  through  the  inter- 
section of  pa  and  ab  as  shown. 
Similarly,  PC  is  the  pressure  on  the 
joint  c,  PD  on  the  joint  d,  etc. 
Thus  it  is  seen  that  the  lines  of 
action  of  the  resultant  pressures  on  the  successive  joints  of  an 
arch  are  the  strings  of  a  funicular  polygon,  the  corresponding 
rays  representing  the  magnitudes  of  these  pressures.  This 
funicular  polygon  will  be  referred  to  as  the  line  of  pressure  of 
the  arch,  although  the  line  of  pressure  or  line  of  resistance,  as 
commonly  defined,  is  the  broken  line  joining  the  centres  0}  pres- 
sure of  the  successive  joints. 

When  the  arch  and  its  loading  are  symmetrical  it  is  evident 
that  the  line  of  pressure  will  also  be  symmetrical  with  reference 
to  a  vertical  through  the  crown,  and  hence  the  pressure  at  the 
crown  will  be  horizontal.  In  this  case  only  one-half  the  arch- 
ring  need  be  considered. 

88.  A  Test  of  Stability.  In  order  to  satisfy  condition  (i)  of 
Art.  85,  in  case  of  rectangular  joints,  it  is  necessary  for  the  re- 
sultant pressure  to  act  within  the  middle  third  of  the  depth  of 
the  joints  (Art.  84) ;  or,  as  commonly  stated,  the  line  of  pressure 
must  lie  within  the  middle  third  of  the  depth  of  the  arch-ring. 
If,  then,  it  is  found  impossible  to  draw  any  funicular  polygon 
for  the  given  system  of  loads  which  will  satisfy  this  condition, 


GRAPHIC  STATICS. 


the  proposed  arch  is  unsatisfactory  and  must  be  altered  in  one 
or  more  of  the  following  particulars:  i.  Thickness  of  arch- ring; 
2.  Form  of  arch- ring;  3.  Distribution  of  loading. 

A  method  of  determining  the  possibility  of  drawing  a  funicular 
polygon  which  will  satisfy  the  above  condition  is  illustrated  in 
case  of  the  segmental  arch  of  Fig.  72. 


a   i    & 


FIG.  72. 

The  arch  is  loaded  symmetrically,  the  loads  being  assumed 
vertical.  The  two  curves  drawn  include  between  them  the 
middle  third  of  the  thickness  of  the  arch-ring.  The  half-arch 
is  divided  into  equal  segments,  the  loads  supported  by  these 
being  AB,  BC,  etc.  The  letters  a,  b,  etc.,  will  be  used  to  represent 
the  strings  of  the  funicular  polygons  and  also  the  joints  of  the 
arch,  a  being  the  joint  at  the  crown  and  g  the  joint  at  the  spring- 
ing. These  so-called  joints  need  not  be  actual  joints  but  only 
convenient  divisions  of  the  arch- ring. 

The  constructions  of  either  Arts.  23,  24,  or  25  may  be  employed. 
Using  first  the  construction  of  Art.  25,  we  proceed  as  follows: 
Assuming  any  pole  P  on  a  horizontal  line  through  A ,  any  funicu- 
lar polygon  XY  is  drawn,  the  strings  being  extended  to  intersect 
the  string  a  at  the  points  bf,  c',  etc.,  as  shown.  These  points  of 
intersection  locate  the  resultant  loads  lying  between  the  corre- 
sponding joints  (see  Art.  25). 

We  now  proceed  to  seek  a  polygon  whose  various  strings  pierce 


MASONRY  ARCHES,  ABUTMENTS,  ETC.  121 

the  corresponding  joints  w'thin  the  limiting  curves.  Judging  from 
the  general  form  of  the  polygon  X  Y,  it  appears  probable,  for  ex- 
ample, that  one  drawn  through  the  points  X  and  Y'  will  satisfy  the 
condition.  The  strings  d,  e,  /,  and  g  of  this  polygon  are  drawn 
as  follows:  Draw  the  string  g  through  Y' ',  its  direction  being 
Y'gf;  then  draw  /,  e,  etc.,  in  turn  so  as  to  pass  through  /',  ef, 
etc.,  as  shown.  It  is  observed  that  this  polygon  falls  outside  the 
required  limits,  its  point  of  greatest  departure  being  at  the  joint 
d.  It  thus  becomes  apparent  that  a  polygon  drawn  through  Yf 
and  touching  the  outer  limiting  curve  at  the  joint  d  is  most  likely 
to  satisfy  the  requirement.  This  polygon  might  have  been  con- 
structed by  the  same  method  as  the  preceding  one,  otherwise 
by  Art.  24  as  follows:  Treating  the  polygon  XY  and  the  one  to 
be  drawn  as  one-half  the  funicular  polygons  for  the  whole  arch, 
the  closing  strings  will  be  horizontal  as  shown.  Since  the  re- 
quired polygon  is  to  pass  through  O',  the  intercepts  made  by 
the  vertical  through  O'  are  RS  and  O'S'  respectively.  The 
services  of  the  new  polygon  and  its  pole  P'  are  then  located  as 
described  in  Art.  24.  This  final  polygon  is  seen  to  satisfy  the 
requirement. 

It  should  be  noted  that  if  this  final  polygon  were  the  true 
line  of  pressure,  the  points,  indicated  by  arrows,  where  the  strings 
cut  the  corresponding  joints  would  be  the  centres  of  pressure  at 
such  joints,  the  corresponding  rays  representing  the  magnitudes 
of  these  pressures. 

"For  the  case  of  non-symmetrical  loading  see  Art.  93. 

The  student  should  be  able  to  check  any  graphical  work 
algebraically,  as  in  case  of  frames.  Thus  let  it  be  required  to 
determine  algebraically  the  magnitude  and  location  of  the  crown 
pressure  H  (Fig.  72),  such  that  the  funicular  polygon  will  pass 
through  O'  and  Y'. 

It  is  known  that  any  string  represents  the  line  of  action  of 


122  GRAPHIC  STATICS. 

the  resultant  external  force  acting  between  the  corresponding 
joint  and  the  crown.  Thus  the  string  d  through  Of  is  the  line 
of  action  of  the  resultant  of  H  and  the  loading  above  the  joint  d. 
The  resultant  moment  of  these  forces  about  O'  must  therefore 
be  zero,  i.e., 

H-y  =  M, (i) 

in  which  M  is  the  moment,  about  O',  of  the  loads  above  the  joint 
d.  Similarly, 

H(y+n)  =  M', .     (2) 

in  which  Mf  is  the  resultant  moment,  about  F',  of  the  loads  above 
the  springing. 

By  solving  (i)  and  (2),  the  values  of  H  and  y  may  be  obtained, 

89.  Relation  between  Line  of  Pressure  and  Form  of  Arch- 
ring.  It  is  evidently  desirable,  to  insure  stability,  that  the  true 
line  of  pressure  coincide,  as  nearly  as  possible,  with  the  centre 
line  of  the  arch-ring. 

If  various  funicular  polygons  be  constructed  for  the  same 
system  of  loads,  changing  the  position  of  the  pole  but  keeping 
it  on  the  same  horizontal  line,  all  such  polygons  will  have  the 
same  general  form;  hence,  since  one  such  polygon  constitutes 
the  true  line  of  pressure,  the  most  suitable  form  of  arch  for  a 
given  system  of  loads  may  be  determined  by  observing  the  general 
form  of  funicular  polygons  for  such  loads.  Thus,  in  Fig.  72, 
the  form  of  the  polygon  XY  or  X'Y'  would  suggest  the  elliptical' 
or  the  multiple-centred  circular  arch  to  be  the  best  forms  for  that 
case.  The  most  suitable  form  for  a  load  uniformly  distributed 
horizontally  would  be  parabolic,  since  the  funicular  polygon  for 
such  loading  is  a  parabola  (Art.  13).  And,  in  general,  an  arch 
comparatively  flat  at  the  crown  is  suitable  when  the  load  in- 
creases in  intensity  from  the  crown  towards  the  abutments, 
while  a  pointed  arch  is  better  adapted  to  the  reverse  condition. 


MASONRY  ARCHES,  ABUTMENTS,  ETC. 


123- 


Conversely,  when  the  form  of  arch  is  given,  its  most  suitable 
distribution  of  loading  can  be  determined  in  a  general  way  by 
treating  its  centre  line  as  a  funicular  polygon  and  determining 
the  corresponding  distribution  of  loads.  The  method  is  illustrated 
in  Fig.  73. 


FIG.  73. 
Let  MN  be  the  centre  line  of  the  arch- ring,  the  short  lines. 

a,  b,  etc.,  being  the  joints  and  the  dotted  lines  indicating  the  lines 
of  action  of  the  loads.     The  strings  of  a  funicular  polygon  to 
coincide  with  MN  will  be  tangent  to  the    curve  at  the  points  a, 

b,  etc.     Therefore,    to    determine    the    distribution    of   loading 
corresponding  to  this  funicular  polygon,  draw  from  any  pole  P 
lines  PA,  PB,  etc.,  parallel  to  tangents  at  a,  b,  etc.     The  lengths 
AB,  BC,  etc.,  thus  determined  indicate  the  relative  magnitudes 
of  the  loads  on  the  divisions  ab,  be,  etc.,  of  the  arch-ring.     This 
distribution  of  loads  may  be  represented  by  laying  off  ordinates 
such  as  ST  =  DE,  etc.     The   area  between  the   curve  XY  thus 
constructed    and    MN  represents   the   distribution  of  loading. 
It  should  be  observed  that  the  diagram  only  represents  the  desired 
distribution  of  loads  in  a  general  way,  since   it  is  not  known  that 
MN  is  the  true  line  of  pressure. 


124  GRAPHIC  STATICS. 

90.  Maximum  and  Minimum  Crown  Pressure.  It  is  evident 
that  as  the  pole  distance  (i.e.,  pressure  at  the  crown)  increases, 
-the  rays  and  strings  of  the  funicular  polygons  become  more 
nearly  horizontal  and  consequently  the  polygons  become  more 
nearly  flat.  Hence,  of  all  polygons  which  can  be  drawn  within 
the  middle  third  (or  other  designated  limits)  of  the  arch-  ring, 
that  one  will  correspond  to  a  maximum  crown  pressure  which 
touches  the  inner  limiting  curve  at  a  point  nearer  the  crown  than 
that  at  which  it  touches  the  outer  curve,  and  conversely. 

Thus,  in  Fig.  72  the  final  polygon  X'Y'  corresponds  to  mini- 
mum crown  pressure  (P'A],  while  the  one  corresponding  to 
maximum  crown  pressure  would  touch  the  inner  limiting  curve 
at  the  crown  and  the  outer  limiting  curve  at  some  point  nearer 
the  springing. 

It  should  be  noted  in  this  connection  that,  other  things  equal, 
as  the  rise  of  an  arch  increases  the  crown  pressure  diminishes. 

It  should  also  be  noted  that  when  three  points  on  the  line  of 
pressure  (corresponding  to  the  three  hinges  of  a  three-hinged 
arch)  are  given,  the  magnitude  and  direction  of  the  resultant 
pressure  at  such  points  can  be  determined  algebraically  in  a 
manner  similar  to  that  explained  in  Art.  52. 

r>  91.  Location  of  the  True  Line  of  Pressure.  It  remains  to 
consider  the  question  as  to  which  of  the  infinite  number  of  funicu- 
lar polygons  corresponding  to  a  given  system  of  arch  loads  con- 
stitutes the  true  line  of  pressure. 

The  prominent  methods  *  which  have  been  suggested  or  are 
in  use  for  locating  the  true  line  of  pressure  may  be  divided  into 
two  classes,  namely: 

I.  Methods  based  on  the  "  Theory  of  the  Elastic  Arch."  2 


1  See  Baker's  "Masonry  Construction  "  for  discussion  of  the  various  methods. 

2  See    Weyrauch's    "Theorie    der  Elastigen   Bogentrager";    also,   Lanza's 
**  Applied  Mechanics." 


MASONRY  ARCHES,  ABUTMENTS,  ETC.  125 

II.  Methods  not  deduced  mathematically,  but  rather  con- 
ventional^  in  character,  the  only  substantial  evidence  of  their 
reliability  being  apparently  derived  from  observation  and  ex- 
perience in  their  use.  These  two  classes  of  solutions  will  be 
discussed  briefly. 

CLASS  I.  It  has  been  frequently  suggested  that  a  masonry 
arch  be  treated  in  the  same  manner  as  a  continuous  iron  arch- 
fixed  at  the  ends,  the  solutions  for  the  latter  case  being  based  on 
the  "Theory  of  the  Elastic  Arch."  These  solutions1  may  be 
designated  as  follows: 

A.  The  exact  solution,  in  which  the  general  formulas  derived 
from  the  "Theory  of  the  Elastic  Arch"  are  employed.     The  line 
of  pressure  thus  located  may  be  taken  to  conform  within  very 
slight  limits  of  error  to  the  theory.     This  solution  is  very  labo- 
rious. 

B.  The  approximate  solution,  in  which  all  but  one  of  the 
terms  of  the  general  formulas  are  omitted,  leaving  them  in  the 
following  form: 


This  method  is  the  one  commonly  presented  in  text-books, 
It  is  adapted  to  graphical  methods  of  solution. 

C.  A  proposition  by  Dr.  Winkler  forms  the  basis  of  a  solu- 
tion which  may  be  properly  included  under  Class  I,  since  it  is 
derived  from  the  formulas  of  solution  B.  This  proposition  is 
as  follows:  "For  an  arch  of  constant  section,  that  line  of  resist- 
ance (pressure)  is  approximately  the  true  one  which  lies  nearest 
the  axis  of  the  arch-  ring  as  determined  by  the  method  of  least 
squares." 

No  approximate    method  should  be  employed  without  first 

1  See  Lanza's  "Applied  Mechanics." 


126  GRAPHIC  STATICS. 

ascertaining  the  amount  of  error  involved.  For  this  purpose  a  com- 
parison of  the  results  to  be  obtained  by  the  use  of  methods  A, 
B,  and  C  was  made  in  the  cases  shown  in  Plate  III.  Two  extreme 
cases  were  taken,  namely:  i.  A  semicircular  arch;  2.  A  segmental 
arch  whose  ratio  of  rise  to  span  is  i  :  10.  In  each  case  three 
systems  of  loads  were  employed,  namely:  i.  A  concentrated 
load  at  the  crown;  2.  A  load  uniformly  distributed  over  the  entire 
span;  3.  A  load  uniformly  distributed  over  the  half-span. 

The  arches  were  of  uniform  thickness  (shown  at  the  crown), 
ratio  of  span  to  thickness  of  arch- ring  being  about  n.6.  The 
lines  of  pressure  (C)  were  located  by  the  calculus,  not  by  trial, 
so  as  to  conform  exactly  .with  Winkler's  proposition. 

The  results  will  not  be  discussed  in  detail,  but  the  following 
conclusions  seem  to  be  warranted. 

1.  Method  C  gives  in  each  case  (except  Case  4)  a  line  of 
pressure  deviating  to  such  an  extent  from  that  of  Method  A  as 
to  prove  the  unreliability  of  the  former. 

2.  Method  B  gives  a  line  of  pressure  agreeing  closely  with  (A) 
in  case  of  the  semicircular  arch,  but  deviating  considerably  from 
the  latter  in  the  case  of  the  segmental  arch.     It  is  evident  that 
Method  B  must  be  used  with  discrimination  if  the  results  are 
to  be  trustworthy.      In  'general,  the    greater   the  ratio  of  span 
to   thickness   of    arch-ring    and    the    greater  the  ratio    of    rise 
to  span,  the  less  will  be  the  difference  between  the  results  obtained 
from  Methods  A  and  B. 

CLASS  II.  Of  these,  the  method  of  "Least  Crown  Pressure" 
seems  to  be  most  commonly  employed.1 

The  theory  of  least  crmvn  pressure  is  essentially  that  the  true 
line  of  pressure  is  that  one  which,  lying  within  the  middle  third 
(or  other  designated  limits)  of  the  arch-ring,  corresponds  to  a 

1  See  Baker's  "  Masonry  Construction." 


MASONRY  ARCHES,  ABUTMENTS,  ETC.  127 

minimum  crown  pressure  (see  Art.  90).  It  appears  to  be  based 
upon  the  observation  that  most  arches  settle  at  the  crown  when 
the  centring  is  removed,  and  upon  the  consideration  that  the 
crown  pressure  is  a  passive  force  developed  by  .the  tendency  of 
the  two  half-rings  to  tip  towards  each  other,  and  is  the  least  force 
that  is  necessary  to  prevent  such  overturning. 

If  the  arch  settles  at  the  crown  (as  a  result  of  rotation,  not 
sliding,  of  the  arch-stones),  the  tendency  would  appear  to  be  to 
open  the  joints  as  shown  in  Fig.  74,  the  centre  of  pressure  moving 


FIG.  74.  FIG.  75. 

upward  at  the  crown  and  inward  at  the  haunches.  The  resulting 
position  of  the  line  of  pressure  would  thus  agree  in  a  general  way 
with  the  case  of  minimum  crown  pressure  (Art.  90).  (It  will  be 
observed  that  all  the  lines  of  pressure  A,  Plate  III,  also  lie  above 
the  centre  of  the  arch- ring  at  the  crown  and  within  it  at  points 
nearer  the  abutments.) 

The  joints  a,  a,  where  the  tendency  to  open  at  the  extrados 
is  greatest,  are  called  the  joints  0}  rupture.  They  correspond 
to  the  points  where  the  line  of  pressure  corresponding  to  mini- 
mum crown  pressure  touches  the  inner  limiting  curve. 

In  the  case  of  a  pointed  arch,  or  an  arch  very  lightly  loaded 
at  the  crown  and  heavily  loaded  at  the  haunches,  the  tendency 
may  be  for  the  crown  to  rise  and  the  haunches  to  move  inward 
(Fig-  75)- 

The  various  other  methods  of  Class  II  differ  from  the  pre- 
ceding only  in  details.  Thus  the  true  line  of  pressure  may  be 


128  GRAPHIC  STATICS. 

assumed  to  be  that  one  which  intersects  the  crown  joint  at  the 
middle  of  its  depth;  at  one- third  the  depth  from  the  extrados, 
etc.;  satisfying,  also,  other  conditions  of  a  similar  character. 
Apparently  all  such  conditions  are  substantially  expressions  of 
judgment  only. 

Comparison  oj  the  Methods  oj  Classes  I  and  II.  The  fact 
that  the  masonry  arch  is  built  up  of  blocks  instead  of  the  material 
being  continuous  and  homogeneous,  as  in  case  of  the  elastic  arch, 
renders  it  questionable  as  to  whether  the  same  methods  of  solu- 
tion are  applicable  to  both.  It  must  also  be  recognized  that  the 
conditions  which  determine  the  location  of  the  line  of  pressure 
of  a  masonry  arch  are  commonly  numerous  and  uncertain. 
Among  such  conditions  are  evidently  the  following: 

1.  The   loading.     This    is    generally    uncertain    in    amount, 
distribution,  and  direction.     In  the    case    of   arches  supporting 
masonry  walls  the  uncertainty  is  particularly  great,  but  it  evidently 
exists  to  some  extent  in  all  cases. 

2.  The  behavior  of  the  arch  as  to  the  kind  and  amount  of 
distortion  occurring  when    the    centring  is  removed  and  subse- 
quently.    Evidently  its  behavior  will   depend   upon   the   quality 
of  the  material  and  workmanship,  the  rigidity  of  the  supports, 
etc.,  as  well  as  upon  the  loading. 

It  is  evident  that  reliable  formulas  cannot  be  derived  from 
uncertain  data,  however  correct  the  reasoning  may  be.  Under 
such  circumstances  the  final  test  of  the  reliability  of  any  method 
of  solution  must  be  agreement  with  the  results  of  experiment 
and  experience.  In  this  respect  the  methods  of  Classes  I  and  II 
stand  on  the  same  footing,  as  each  requires  experimental  veri- 
fication. 

For  the  present,  in  lieu  of  adequate  experimental  results  we 
must  rely  largely  upon  the  experience  of  the  past  in  the  construc- 
tion of  masonry  arches;  making  use  of  that  line  of  pressure 


MASONRY  ARCHES,  ABUTMENTS,  ETC.  129^ 

which  appears  to  be  most  reliable  and  seeing  to  it  that,  with  such 
line  of  pressure,  the  conditions  of  Art.  85  are  fulfilled. 

The  author  has  no  recommendation  to  make  as  regards  the 
best  method  of  locating  the  line  of  pressure  of  a  masonry  arch. 
In  the  following  examples,  however,  the  condition  that  "the  true 
line  of  pressure  is  that  one  which,  lying  within  the  middle  third, 
of  the  arch-ring,  corresponds  to  the  minimum  crown  pressure," 
will  be  employed;  the  portion  of  the  solution  subsequent  to  the 
location  of  the  line  of  pressure  being  the  same,  however  this 
line  may  be  located. 

In  order  to  fix  the  position  of  the  line  of  pressure,  three  hinges 
have  been  employed  to  some  extent  in  Germany  and  elsewhere. 
The  line  of  pressure  is  thus  determined  by  the  condition  that  it 
must  pass  through  the  hinges,  the  same  as  in  case  of  the 
three-hinged  iron  arch. 

92.  Example.  Figure  i,  Plate  III,  is  one-half  of  a  sym- 
metrical full-centred  arch  in  a  masonry  wall  whose  height  is  limited 
by  a  horizontal  line,  as  shown.  It  is  required  to  draw  the  line  of 
pressure  according  to  the  theory  of  least  crown  pressure,  and  to 
determine  if  the  arch  satisfies  the  conditions  of  stability. 

The  half-ring  is  divided  by  radial  lines,  which  need  not  coin- 
cide with  the  actual  joints  of  the  arch.  It  will  be  assumed  that 
each  of  these  divisions  supports  the  weight  of  the  portion  of  wall 
directly  above  it,  as  indicated  by  the  vertical  lines.  If  the  specific 
gravity  of  the  material  above  the  arch- ring  is  the  same  as  that 
of  the  arch-ring,  the  load  supported  by  any  single  division,  as 
mn,  is  proportional  to  the  area  of  the  polygon  mnn'm'.  If  the 
specific  gravities  are  unequal,  the  vertical  ordinates  may  be 
altered  in  length  so  that  the  areas  above  the  divisions  will  repre- 
sent weights  to  the  same  scale  as  the  areas  of  the  divisions  them- 
selves. Otherwise  the  weights  of  the  divisions  and  material 


I3o 


GRAPHIC  STATICS. 


TABLE  or  LOADS. 
(Fie.  i,  PLATE  III.) 


above  them  may  be  dealt  with  separately.  In  this  example,  the 
wall  is  of  uniform  thickness  and  the  weight  of  the  masonry  is, 
throughout,  160  Ibs.  per  cubic  foot. 

Considering  one  foot  thickness  of  wall,  the  loads  (see  table) 
are  calculated  by  multiplying  the  cor- 
responding areas  by  160  Ibs.  HI  is 
taken  to  include  two  divisions  to 
avoid  confusing  the  drawing;  //  is 
the  weight  of  the  masonry  to  the 
right  of  the  line  i;  JK  is  the  weight 
of  the  masonry  below  RS. 

The  resultant  loads  act  at  the 
centres  of  gravity  of  the  areas.  The 
centre  of  gravity  of  the  division  mn 
is  O',  and  the  centre  of  gravity  of  the 
trapezoid  above  this  division  is  O" 
(see  Art.  79).  The  centre  of  gravity 
O  of  the  entire  area  mnn'm'  is  then  found  by  dividing  the  line 
O'O"  into  parts  inversely  proportional  to  these  areas.  The  centres 
of  gravity  are  indicated  by  circles. 

The  line  of  pressure  is  now  constructed  as  follows:  The  loads 
AB,  BC,  etc.,  are  plotted  to  scale,  and,  selecting  any  pole  P  on  a 
horizontal  line  through  A,  a  preliminary  polygon  xy  is  drawn,  the 
point  x  being  one-third  the  depth  of  the  joint  below  the  extrados. 
In  drawing  this  polygon,  the  intersections  cf,  df,  etc.,  of  its  various 
strings  with  the  string  a  are  marked  (see  Art.  88). 

The  line  of  pressure  desired  is  such  that  the  resultant  pressure 
at  the  joint  of  rupture  will  act  at  one-third  the  depth  of  the  joint 
from  the  intrados.  The  joint  of  rupture  may  be  determined  by 
trial  as  follows:  Judging  that  it  will  be  near  the  joint  g,  we  trisect 
that  joint  at  i  and  draw  the  string  g  through  i  and  gf.  The 
adjacent  strings  (not  shown  in  the  drawing)  are  then  drawn,  and 


Weight 

Weight 

Division. 

of 

above 

Division. 

Division. 

Lbs. 

Lbs. 

AB 

540 

1360 

BC 

540 

1440 

CD 

540 

1570 

DE 

540 

1700 

EF 

540 

1790 

FG 

540 

1760 

GH 

540 

1540 

HI 

1080 

1600 

IJ 

— 

10240 

JK 

7680 

MASONRY  ARCHES,  ABUTMENTS,  ETC,  131 

it  is  found  that  the  string  /  falls  farthest  outside  the  middle-third 
limit.  This  joint  is  then  trisected  at  2  and  the  string  2]'  of  a  new 
polygon  is  drawn.  This  polygon  is  completed  by  drawing  the 
remaining  strings  in  succession  through  e',  df,  etc.,  and  is  found 
to  satisfy  the  conditions  except  near  the  springing  where  it  falls 
outside  the  middle-third  limit.  This  portion  of  the  arch,  however, 
can  be  treated  as  part  of  the  abutment.  The  centres  of  pressure 
at  the  different  joints  are  indicated  by  arrows. 

Aside  from  the  condition  that  the  pressure  must  act  within 
the  middle  third  of  the  arch-ring,  the  resistance  to  sliding  and 
crushing  must  be  investigated. 

As  regards  sliding,  it  is  seen  that  the  direction  of  the  resultant 
pressure  at  each  joint  is  very  nearly  normal  to  the  joint,  with  the 
exception  of  the  springing  plane,  where  the  pressure  p'i  makes 
with  the  normal  an  angle  greater  than  tan"1^.  When,  however, 
the  weight  //  is  combined  with  P'I,  the  resultant  pressure,  P"J 
(Art.  96),  satisfies  the  requirement  for  safety  against  sliding. 

For  crushing,  the  maximum  compression  stress  is  to  be  cal- 
culated at  each  dangerous  joint.  For  example,  the  resultant 
stress  on  the  joint  /,  found  by  scaling  off  the  ray  P'F,  is  1310x3 
Ibs.  The  area  of  the  surface  of  the  joint  is  iXif=if  sq.  ft.  = 
252  sq.  in.  Average  pressure  per  sq.  in.  = -if -^.=  52  Ibs.  Hence 
the  maximum  stress  =  2X52  =  104  Ibs.  per  sq.  in.  (see  Art.  84). 
This  stress  must  not  exceed  the  working  compression  strength 
of  the  masonry.  (For  strength  of  masonry  see  Lanza's  "Applied 
Mechanics  "  and  other  references.) 

93.  Unsymmetrical  Cases.  When  the  arch  or  loading  is 
unsymmetrical,  the  line  of  pressure  is  also  unsymmetrical,  and 
must  therefore  be  drawn  for  the  whole  arch.  The  construction 
of  the  line  of  pressure  involves  the  problem  of  drawing  a  funicular 
polygon  through  three  points  (see  Arts.  23  and  24). 


I32 


GRAPHIC  STATICS. 


EXAMPLE.  (Fig.  76.)  Given  a  segmental  arch  of  16  ft.  span 
and  3  ft.  rise.  Thickness  of  arch- ring  =  i  \  ft.  The  left  and  right 
halves  are  loaded  with  3200  Ibs.  and  6400  Ibs.  respectively,  per 
foot  width  of  arch,  these  loads  being  uniformly  distributed  over 
the  arch-ring.  It  is  required  to  construct,  if  possible,  the  line  of 
pressure  in  accordance  with  the  method  of  "least  crown  pressure." 


a!  6 


FIG.  76. 

The  arch-ring  is  divided  into  sixteen  equal  divisions,  and  the 
load  supported  by  each  is  assumed  to  act  at  the  middle  of  its  outer 
surface.  The  loads  are  plotted  to  scale,  and  the  funicular  polygon 
xy  is  constructed,  using  P  for  pole.  Selecting  the  points  t' ',  i/, 
and  ur  at  one-third  and  two-thirds  the  depth  of  the  joints  from 


MASONRY  ARCHES,  ABUTMENTS,  ETC.  133 

the  intrados,  the  funicular  polygon  which  will  pass  through  these 
points  is  located  as  explained  in  Art.  24  (see  Fig.  76  A).  This 
polygon  is  not  drawn,  but  the  points  on  it,  falling  outside  the 
middle-third  limits,  which  serve  to  locate  the  final  line  of  pres- 
sure, are  marked  by  dots. 

It  is  seen  that  to  the  right  of  the  crown  this  polygon  rises  above 
the  middle-third  limit,  while  near  the  left  abutment  it  falls  below. 
From  the  position  of  these  points  it  appears  probable  that  a  polygon 
drawn  through  the  three  points  i',  2',  and  vf  will  fall  within  the 
specified  limits.  The  vertices  of  this  polygon  (drawn  in  full  lines) 
were  located  in  the  same  manner  as  the  preceding  one  (see  Fig. 
76  B),  and  it  is  seen  that  the  centres  of  pressure  at  all  the  joints 
fall  within  the  required  limits. 

The  pole  P'  of  this  final  polygon  was  located  by  the  method 
of  Art.  23,  i.e.,  PZ  and  PZ'  were  drawn  parallel  to  i,  2  and  2,  v 
respectively;  then  ZP'  and  Z'P'  were  drawn  through  Z  and  Z' 
parallel  to  i',  2'  and  2',  if  respectively,  thus  locating  P' .  This 
pole  might  also  have  been  located  by  Art.  24.  The  resultant 
compression  at  the  various  arch  sections  is  represented  by  the 
rays  of  this  final  polygon. 

The  stability  of  the  abutments  is  essential  to  that  of  the  arch 
(see  §  3). 

§  3.  Abutments,  Piers,  etc. 

94.  Conditions  of  Stability.      The  general  conditions  of  sta- 
bility of  §  i  are  applicable  to  any  piles  of  masonry  subjected  to 
the  action  of  external  forces,  such  as  the  thrust  of  an  arch  or 
truss,  pressure  of  earth,  water,  wind,  etc. 

95.  Example  i.      Fig.    77    is   an   abutment   subjected   to   a 
horizontal  pressure  AB  and  a  vertical  pressure  BC,  their  re- 


5  54 


GRAPHIC  STATICS. 


.  77 


sultant.  AC}  acting  at  the  point  c.     CD,  DE,  EF}  and  FG  are 

the  weights  of  the  four  divisons 
of  the  masonry,  their  centres  of 
gravity  being  marked  by  circles. 
The  pressure  on  the  joint  d  is 
the  resultant  of  AC  and  the 
weight  CD  of  the  first  block. 
This  resultant  is  AD,  its  line  of 
action  passing  through  the  inter- 
section of  ac  and  a  vertical 
through  the  centre  of  gravity  of 
the  block.  The  point  of  appli- 
cation of  this  resultant  pressure, 
at  d,  is  indicated  by  an  arrow.  The  resultant  pressures  at  the 
remaining  joints  are  found  in  a  similar  manner. 

The  conditions  of  safety  as  regards  sliding,  overturning, 
and  crushing  have  been  previously  discussed.  The  maximum 
pressure  on  the  soil  must  also  be  kept  within  safe  limits  (see  Baker's 
"Masonry  Construction,"  Chap.  X). 

A  broken  line  connecting  the  points  of  application  c,  d,  e,  /, 
and  g  of  the  resultant  pressures  on  the  successive  joints  of  an 
abutment  is  commonly  called  the  line  of  resistance  or  line  oj  pres- 
sure, as  in  case  of  an  arch  (see  Art.  87). 

96.  Example  2.  The  abutments  of  a  masonry  arch  can  be 
considered  in  connection  with  the  arch.  In  Fig.  i,  Plate  III,  the 
weights  //  and  JK,  lying  above  and  below  the  joint  RS,  are 
plotted  to  half  scale  (4000  Ibs.  =  i  in.),  the  pole  for  these  two  loads 
being  located  by  bisecting  the  ray  P'l  at  P" .  The  resultant  pres- 
sure on  RS  is  P"J)  its  line  of  action  p"j  passing  through  the  inter- 
section of  pfi  and  ij.  The  resultant  pressure  on  the  base  is  P"K, 
its  line  of  action  being  p"k.  To  find  the  maximum  intensity  of 


MASONRY  ARCHES,  ABUTMENTS,  ETC. 


135 


the  pressure  on  the  base  we  have  F"K  =  g.i  (inches)  X  4000  = 
36400  Ibs.  By  measurement,  p"k  acts  i.i  ft.  from  the  centre  of 
the  base.  The  bearing  area  is  7fXi  =  7l  sq-  ft-  Substituting 

these  values  in  the  formula  /=     +     ^»  we  have 


36400       36400.1.1 

7-75      TVi-(7-75)3 


ibs.  per  Sq.  ft. 
60  Ibs.  per  sq.  inch. 


97.  Example  3.     Let  Fig.  78  represent  a  pier  supporting  the 
thrust  of  an  arch  on  each  side. 

These  thrusts  are  AB  and  BC, 

their  point  of  intersection  O  lying 

on   the    centre    line   of   the   pier. 

Let  CD,  DE,  EF,  and  FG  repre- 

sent    the    weights     of    the    pier 

divisions,    CD    being   the   weight 

of   the    pier   masonry    above   the 

joint   d.     The   resultant   pressure 

on  d  is  AD,  its  line  of  action  Od 

passing    through    O.     The    lines 

of  action  of  the  pressures  on  the 

remaining    joints    will    also    pass 

through     O,     since     the     vertical 

through  this  point  contains  the  centres  of  gravity  of  all  the  blocks. 

These  lines  are  Od,  Oe,  Of,  and  Og,  drawn  parallel  respectively 

to  AD,  AE,  AF,  and  AG.     If  the  thrusts  of  the  two  arches  are 

equal  and  equally  inclined,  the  resultant  pressure  on  the  pier  will 

evidently  be  vertical. 

98.  Example  4.     Let  Fig.  79  represent  a  chimney  subjected 
to  wind  pressure.     The  weights  of  the  portions  ab,  be,  etc.,  are 


FIG.  78. 


-«36  GRAPHIC  STATICS. 

AB,  BC,  etc.,  and  the  wind  pressures  on  these  portions  are  AB', 
3'Cf,  etc.  The  lines  of  action  of  these  wind  pressures  are  the 
horizontal  dotted  lines  ab,  be,  etc.  With  any  pole  P  draw  the 
funicular  polygon  mn  for  the  wind  loads.  (For  this  purpose 
these  loads  should  be  plotted  to  a  larger  scale.) 


F'ED'CB 


FIG.  79. 

To  find  the  resultant  pressure  at  any  section,  as  e,  the  line  of 
action  tr  of  the  resultant  wind  pressure  above  that  section  is 
located  by  the  intersection  of  the  strings  a  and  e. 

The  pressure  at  the  section  e  is  the  resultant  of  this  wind 
pressure  and  the  weight  of  masonry  above  e.  The  line  of  action 
rs  of  this  pressure  will  act  through  r,  the  point  of  intersection  of 
the  resultant  wind  pressure  and  weight  of  masonry  above  e,  and 
its  direction  will  be  parallel  to  E'E,  its  magnitude  being  repre- 


MASONRY  ARCHES,  ABUTMENTS,  ETC.  137 

sented  by  the  length  of  E'E.     The  resultant  pressure  at  any  other 
section  can  be  determined  in  a  similar  manner. 

The  preceding  examples  will  serve  to  indicate  the  method  of 
determining  the  stability  of  such  structures  when  the  loads  are 
known. 


-17000  —17000  —  HCCO 

-45200  —43200  —  3SSOO 


Fig.  1C 


ELA.TE  1. 


Fig.  2 


TABLE  OF  STRESSES.      DEAD  LOAD  WITH  WIND  ON  LEFT  SIDE. 


UPPER  CHORD 

LOWER  CHOR 

3    VERTICALS 

DIAGONALS 

••'.;>:•;  TCR 

CZJCG300      C 

1 

ISiU)      ' 

Biiies^o   c 

1 

OC.V'J    T 

AK 

9iW     C 

CY  *W;0     C 

X 

X 

SHOO    C 

Yxtmjjo  • 

'  B 

=W,'8«CO     C 

V    GS4CO 

wi  2isco    ; 

vvi-JuToy 

:  ^ 

1031)0     C 

FUfWCiiO      C 

Tj  42700 

y 

ssoi-o    ; 

'Tji'.meo  ' 

1] 

6300     C 

GS   773tJl)     C 

R 

i27t,U     T 

1 

30UO    t 

-'.'  i  ','.••..,    T 

llfl 

0 

HO   i<8w     C 

iiSM)     T 

•-,. 

Hit,;,'    C 

I  ? 

Gl'OO    T 

6F|         0 

10    7*t«.>0      C 

\ 

i^ioo    T  1  PO  i  icsoo  c 

N 

iul'.J    T 

FEl    WCO    T 

JM   193CO     G 

L 

22500     C  UM 

31SCO    C 

L 

oOit:.)    T 

ILK 

iSSCO     C 

FGH 


Fig.  1  2000  Tos. 


F\g.2 


PLATE  II 


PLATE  II.I 


2.  LOAD  UNIFORMLY 
DISTRIBUTED  OVER  SPAN 

(LINES  OF  PRESSURE,  METHODS  B  AND  C, 
COINCIDE  CLOSELY  WITH  AXIS  OF  ARCH  RING) 


4.  LOAD  UNIFORMLY  DISTRIBUTED  OVER  SPAN 


LOAD  UNIFORMLY  DISTRIBUTED  OVER  LEFT  HALF-SPAN 


ARCHES  OF  UNIFORM  SECTION.  FIXED  ENDS* 

(  METHOD  A FULL  LINES 

LINES  OF  PRESSURE      1  METHOD  B BROKEN  LINES 

j.    (  METHOD  C DOTS  AND  DASHES 


6.   LOAD  UNIFORMLY  DISTRIBUTED  OVER  LEFT  HALF-SPAN. 


T& 

,  THE  LIBRARY 

2  'U  UNIVERSITY  OF  CALIFORNIA 

Santa  Barbara 


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